TRIG: Inverse Trigonometric Functions

00:58:05
https://www.youtube.com/watch?v=b-d91UK0_3M

Resumen

TLDRThe video lecture is designed to assist students at the University of South Florida with understanding inverse trigonometric functions, specifically section 4.6 of their course material. It covers fundamental concepts of inverse functions including their domains, ranges, and how they can be derived from standard trigonometric functions by domain restrictions. The lecturer explains the horizontal line test, which ensures a function is one-to-one hence invertible, and how to work with angles using properties and identities of inverse functions. Steps for solving various problems are demonstrated, such as calculating inverse trigonometric function values, understanding when an expression exists, and handling special cases using diagrams and algebra. The lecture advises students to use their calculator features for inverse calculations and to refer back to provided formulas and diagrams for a comprehensive understanding.

Para llevar

  • 🎓 The lecture complements the classroom learning for USF students.
  • 📚 Focus is on inverse trigonometric functions and key properties.
  • 👩‍🏫 Includes background information for standalone learning.
  • 🔄 Emphasizes understanding domains and ranges of arc functions.
  • 📝 Stresses the importance of the horizontal line test for invertibility.
  • 🔍 Diagrams and sketches help visualize inverse function properties.
  • 🔢 Demonstrates problem-solving with inverse trigonometric values.
  • 🎯 Points out the need for domain checking to avoid undefined results.
  • 💡 Highlights reciprocal identities for solving secant and cosecant.
  • 📏 Advises on using calculator functions for practical calculations.

Cronología

  • 00:00:00 - 00:05:00

    The lecturer introduces the topic of inverse trigonometric functions, emphasizing that it serves as supplementary material for students from the University of South Florida for their section 4.6. The lecture aims to cover all necessary information on inverse trig functions to prepare students for upcoming exams, including those outside the class who want to learn independently. The lecturer provides a brief overview of the lecture structure, including background information and examples, while also making the session as comprehensive as possible despite previous recording challenges.

  • 00:05:00 - 00:10:00

    A diagram illustrates the concept of functions and their inverses, emphasizing the opposite nature of their domain and range. Using a metaphor involving crimes and detectives, the lecturer explains the relationship between a function and its inverse. The person committing a crime equates to the original function, whereas the detective represents the inverse function. The lecturer further elaborates on domain-range reversals and key equations that demonstrate how an inverse function allows one to revert outputs to their original inputs.

  • 00:10:00 - 00:15:00

    The horizontal line test is introduced as a criterion for determining if a function has an inverse. The lecturer uses a straight line and a parabola to illustrate this test, explaining that a function must intersect any horizontal line at no more than one point to be invertible. The challenge with trigonometric functions, which are usually periodic and would thus fail this test, is addressed. Solutions involve restricting the domain of trig functions to make them invertible, as will be shown for main trig functions in the next section.

  • 00:15:00 - 00:20:00

    The section explains how trig functions are restricted to define their inverses, starting with sine. For sine, the domain is limited to -π/2 to π/2 to create a one-to-one function suitable for defining arcsine. Similar restrictions are applied to cosine (0 to π) and tangent (-π/2 to π/2) to define their respective inverse functions, with explanations on domain and range specifics for each. The lecture briefly touches on approaches for handling inverse cosecant and secant by using the knowledge of reciprocal identities of sine and cosine.

  • 00:20:00 - 00:25:00

    Using sketches, cosecant and secant functions are illustrated as reciprocals of sine and cosine, leading to their distinct vertical asymptotes and domain restrictions. The range for arccosecant and arcsecant is adapted accordingly. Similarly, for cotangent, its behavior as decreasing over full cycles sets its domain for arc functions. The lecturer stresses the importance of understanding the reciprocal nature of trig functions to work effectively with their inverses, followed by a preparation into practical example solving.

  • 00:25:00 - 00:30:00

    The lecturer presents major equations for each inverse trigonometric function with emphasis on understanding rather than memorization. Differences in presentation include occasionally using color to highlight specific functional properties, like negation impacts, and the equivalences of inputs and outputs between functions and their inverses. The importance of knowing specific domains and ranges for working with inverse functions is highlighted, with reminders on checking that inputs fall within valid ranges before solving examples.

  • 00:30:00 - 00:35:00

    Examples of arcsine values are calculated, focusing first on positive then negative, demonstrating how to maneuver within restrictions to determine angles in different quadrants corresponding to given values. The lecture emphasizes the importance of understanding reference angles and ranges to ensure the correctness of derived angles. It uses errors made without proper domain considerations as teaching moments by checking calculated vs. theoretical values, ensuring proper inverse function logic is consistently applied.

  • 00:35:00 - 00:40:00

    An example using arccosine with positive and negative values shows the importance of recognizing angle orientation in specific quadrants (quadrant 1 for positives, quadrant 2 for negatives). This involves calculating exact reference angles and recognizing how they frame the function's valid range. The lecturer explains via drawing sketches and reflections across axes to assist in visual understanding, linking back to theoretical principles discussed earlier, ensuring students can apply these steps across different inverse trig scenarios.

  • 00:40:00 - 00:45:00

    An example for arctangent demonstrates using domain checks and reciprocal identities. The method involves recognizing angle properties within unit circle sections, using sketches to guide through quadrant identification for sign correction. The exercise stresses recognizing symmetry across the unit circle to aid in solving these equations correctly. The idea is to solidify student confidence in applying inverse calculations in contexts where exact quadrant determination significantly affects outcome validity.

  • 00:45:00 - 00:50:00

    For reciprocal functions like arcsecant, the lecturer demonstrates using cosine because of secant's reciprocal nature. This simplification aids problem-solving as cosine's more intuitive properties are used to derive results effectively, emphasizing sketching and remembering unit circle values. An exercise switches to arccosecant, reaffirming reciprocal function solutions with a similar approach. These sketches show utility and effectiveness in solving trigonometric problems by linking inverse trig understanding to clear visual strategies.

  • 00:50:00 - 00:58:05

    The video ends with examples focusing on inverse trigonometric problem-solving steps, emphasizing the usefulness of pictorial representation for clarity, especially in multi-step inverse calculations. Several problems show applying inverse function formulas appropriately, including noting when expected conditions aren't met, resulting in no solution. Emphasis is placed on continual practice to ensure students can independently navigate inverse functions assignments with correct theoretical and practical understanding.

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Mapa mental

Mind Map

Preguntas frecuentes

  • What is the range for inverse sine function?

    The range for the inverse sine function is from -π/2 to π/2.

  • What does it mean when a function passes the horizontal line test?

    It indicates that the function has an inverse, as each horizontal line intersects the graph at most once.

  • How can you restrict trigonometric functions to have an inverse?

    By limiting their domains to one period where they pass the horizontal line test.

  • What is the domain of the inverse tangent function?

    The domain for the inverse tangent function is all real numbers.

  • What does an inverse function do?

    An inverse function reverses the operation of a function, taking outputs back to the original inputs.

  • Why is the horizontal line test important for inverse functions?

    It's important because a function must be one-to-one to have an inverse, which is ensured by the horizontal line test.

  • What happens when you reflect a trigonometric angle across the x-axis?

    Reflecting across the x-axis changes the sign of the y-coordinate while keeping the x-coordinate the same.

  • How do reciprocal trigonometric functions like cosecant and secant affect angles?

    They require using reciprocal identities to translate problems back into sine or cosine, which are easier to evaluate.

  • Why might an expression involving cosine inverse not have a solution?

    If the input is outside the range [-1, 1], it doesn't satisfy the domain requirements of the inverse cosine function.

  • What basic property can be used to find the angle with inverse sine or cosine?

    The property that sin(arcsin(x)) = x or cos(arccos(x)) = x if x is within the respective function's domain.

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Desplazamiento automático:
  • 00:00:01
    hello everyone and welcome to this video
  • 00:00:03
    lecture on inverse trigonometric
  • 00:00:06
    functions now this video is specifically
  • 00:00:09
    pitched at my students at the University
  • 00:00:11
    of South Florida we were working on
  • 00:00:12
    inverse trig functions this past week
  • 00:00:14
    which would be section 4.6 in our notes
  • 00:00:17
    and we got some of it done but not all
  • 00:00:19
    of it so I'm making this video as a
  • 00:00:21
    supplementary lecture to make sure we
  • 00:00:23
    get through all the information and that
  • 00:00:25
    you are all prepared for upcoming exams
  • 00:00:28
    now if you happen to be a past by
  • 00:00:30
    somebody who saw the video and thought
  • 00:00:32
    I'd like to learn more about inverse
  • 00:00:33
    trick functions don't worry we are going
  • 00:00:36
    to be providing enough background
  • 00:00:38
    information that this video should act
  • 00:00:40
    like a standalone lecture and as long as
  • 00:00:42
    you follow along with me you take notes
  • 00:00:45
    you actually try to work through the
  • 00:00:46
    examples you should walk away with a
  • 00:00:48
    decent understanding of what's going on
  • 00:00:51
    however from what I've said there are
  • 00:00:53
    some things that should be known upfront
  • 00:00:56
    first of all I must really like you USF
  • 00:00:58
    students because I think I've reach done
  • 00:01:00
    this video about 73 billion times that's
  • 00:01:03
    the nature of recording videos that's
  • 00:01:05
    just how it is and this time around I
  • 00:01:07
    think I'm probably going to record it
  • 00:01:09
    and post it warts and all so if you hear
  • 00:01:11
    me say something especially funny you
  • 00:01:13
    know why the second thing you should
  • 00:01:15
    know is since this is going to be kind
  • 00:01:17
    of like a standalone video I am going to
  • 00:01:20
    put in some background information
  • 00:01:22
    however because this is going to be
  • 00:01:24
    primarily supplementary lecture material
  • 00:01:27
    for my students what that means is the
  • 00:01:30
    background material is going to be gone
  • 00:01:32
    through pretty quickly for example I
  • 00:01:35
    have some reminders on this page and I
  • 00:01:37
    am going to go through those reminders
  • 00:01:39
    however they're going to be pretty fast
  • 00:01:41
    the next couple of pages are going to
  • 00:01:43
    have some reminders and some formulas
  • 00:01:44
    that they should already have seen and
  • 00:01:46
    should already know and we're just going
  • 00:01:48
    to pass through them swiftly so we can
  • 00:01:50
    move to the examples which are going to
  • 00:01:52
    be our primary objective with that said
  • 00:01:55
    make sure that you have something to
  • 00:01:57
    write with next to you make sure that
  • 00:01:59
    you have your notes or a good textbook
  • 00:02:00
    or that you are very comfortable drawing
  • 00:02:02
    the original Sixt trig functions and
  • 00:02:05
    let's go ahead and
  • 00:02:06
    begin whenever I start talking about
  • 00:02:09
    inverse functions of any kind I like to
  • 00:02:11
    throw this diagram up on the screen it
  • 00:02:14
    does not actually serve as any kind of
  • 00:02:16
    definition but it is a nice cartoon that
  • 00:02:18
    kind of helps to encapsulate a lot of
  • 00:02:21
    information and I'm going to go through
  • 00:02:24
    that information quickly now so if you
  • 00:02:27
    have a function like the one we've shown
  • 00:02:29
    here f which has an inverse which we
  • 00:02:32
    would call F inverse which is what we've
  • 00:02:34
    written here when we write f with
  • 00:02:36
    negative one like an exponent it does
  • 00:02:38
    not mean a reciprocal then what should
  • 00:02:41
    happen is we should find that F takes
  • 00:02:43
    inputs X in a set a to outputs Y in a
  • 00:02:49
    set B and F inverse should do exactly
  • 00:02:52
    the opposite if you want to think of
  • 00:02:54
    these using real world examples one that
  • 00:02:56
    is near and dear to my heart is the
  • 00:02:58
    example of crimes and detectives don't
  • 00:03:01
    ask me why that is what comes to mind
  • 00:03:03
    but it fits very nicely you see X the
  • 00:03:06
    input in that case would be kind of like
  • 00:03:09
    the culprit who is going to commit the
  • 00:03:11
    crime the opportunity to commit the
  • 00:03:12
    crime the tools all the things that go
  • 00:03:15
    into actually doing the crime before
  • 00:03:16
    it's done the function f takes all of
  • 00:03:19
    those pieces parts and actually commits
  • 00:03:22
    the crime and why is the result of the
  • 00:03:24
    crime the detective then is played by F
  • 00:03:28
    inverse and it starts at the crime
  • 00:03:30
    itself what has happened what the clues
  • 00:03:33
    are surrounding the crime and attempts
  • 00:03:35
    to deduce from that information who the
  • 00:03:38
    culprit was what the means motive and
  • 00:03:40
    opportunity were Etc that is the
  • 00:03:42
    relationship between a function and its
  • 00:03:44
    inverse as pictured here from this
  • 00:03:47
    diagram then we have several things that
  • 00:03:49
    we should note down they are shown here
  • 00:03:51
    but we need to say them a little bit
  • 00:03:52
    more carefully the first concerns domain
  • 00:03:55
    range the function f has a domain of a
  • 00:03:58
    and a range of B meaning that all the
  • 00:04:01
    inputs for f belong to the set a and all
  • 00:04:04
    the outputs for f belong to the set b as
  • 00:04:06
    you can see here from our diagram our
  • 00:04:09
    diagram also says now when you look at
  • 00:04:11
    it that the domain for f inverse ought
  • 00:04:13
    to be B the things that we plug into F
  • 00:04:17
    inverse all sit in the set B and the
  • 00:04:20
    things that come out of f inverse all
  • 00:04:22
    belong to the set a so that means the
  • 00:04:24
    range for f inverses a which you'll
  • 00:04:26
    notice is the reverse of what we had
  • 00:04:29
    before that's very natural that's
  • 00:04:31
    exactly how inverses and functions
  • 00:04:34
    work another thing that we can learn
  • 00:04:36
    from looking at the diagram but which we
  • 00:04:38
    have to be careful about is some
  • 00:04:40
    equation properties of these functions
  • 00:04:43
    the first one on the left hand side says
  • 00:04:45
    if I feed an output y to F inverse I'm
  • 00:04:48
    going to get an input for my function f
  • 00:04:51
    if I feed that input to F the result is
  • 00:04:53
    going to be y again just like what I
  • 00:04:56
    plugged into F inverse similarly if I
  • 00:04:59
    plug the input X into F I'm going to get
  • 00:05:03
    an output if I feed that output to F
  • 00:05:05
    inverse I should come back to the
  • 00:05:07
    original input
  • 00:05:09
    X now those two equations really mean
  • 00:05:12
    follow the arrows in my diagram if I
  • 00:05:15
    start at y feed it to F inverse and then
  • 00:05:17
    feed that to F I come back to Y again
  • 00:05:20
    exactly like my equation said if I start
  • 00:05:23
    at X feed it to F and then feed that to
  • 00:05:27
    F inverse I come back to X again so
  • 00:05:30
    we've kind of already got it in the
  • 00:05:31
    picture another good reason to have the
  • 00:05:33
    picture hanging around it says the same
  • 00:05:35
    thing but it looks nicer the other
  • 00:05:38
    equations we're interested in on the
  • 00:05:39
    right hand side are what we would call
  • 00:05:41
    equivalent equations meaning they mean
  • 00:05:44
    exactly the same thing f ofx = y and F
  • 00:05:48
    inverse of yal X are exactly the same
  • 00:05:50
    information and we've noted that with
  • 00:05:52
    this double-ended arrow in between them
  • 00:05:56
    if you want to think of it this way they
  • 00:05:57
    are just different perspectives on the
  • 00:05:59
    same information if I'm talking about
  • 00:06:01
    myself I might call myself Dr forest or
  • 00:06:04
    Zach if somebody else was talking about
  • 00:06:07
    me they might say he is Zach or they
  • 00:06:09
    might say he is Dr Forest same
  • 00:06:11
    information but from a different
  • 00:06:12
    perspective and that's what these
  • 00:06:14
    equations say whether you come from the
  • 00:06:16
    perspective of f or the perspective of f
  • 00:06:18
    inverse this is the same
  • 00:06:20
    information the final thing that we need
  • 00:06:23
    to include is a function f which has an
  • 00:06:27
    inverse must pass the horizontal line
  • 00:06:29
    test now what does that look
  • 00:06:33
    like well if I take a couple of basic
  • 00:06:36
    functions you can see one of them is a
  • 00:06:38
    straight line and one of them is a
  • 00:06:40
    parabola a function that passes the
  • 00:06:43
    horizontal line test should see the
  • 00:06:45
    following if I draw any horizontal
  • 00:06:49
    line it should do one of two things
  • 00:06:52
    either it should fail to intersect the
  • 00:06:54
    graph of my function entirely or like
  • 00:06:58
    you see in my picture here it should
  • 00:07:00
    intersect the graph of my function
  • 00:07:02
    precisely
  • 00:07:03
    once this must happen for every single
  • 00:07:06
    horizontal line I
  • 00:07:08
    draw as you can see with a straight line
  • 00:07:11
    this works every horizontal line I draw
  • 00:07:13
    passes through my line precisely once
  • 00:07:15
    and that's okay if you look at my
  • 00:07:18
    Parabola though we're not so lucky yes
  • 00:07:20
    there are some horizontal lines that
  • 00:07:22
    don't pass through the graph of my
  • 00:07:24
    Parabola yes there is a horizontal line
  • 00:07:27
    it's the x-axis that only touches my
  • 00:07:29
    parab Parabola
  • 00:07:30
    once
  • 00:07:32
    however because I can draw a horizontal
  • 00:07:35
    line just one that does not intersect
  • 00:07:39
    once it intersects
  • 00:07:41
    twice that means that my Parabola does
  • 00:07:44
    not pass the horizontal line test from
  • 00:07:46
    that from this horizontal line test you
  • 00:07:49
    can usually learn some
  • 00:07:50
    information as we said an inverse a
  • 00:07:54
    function that has an inverse must be
  • 00:07:55
    able to pass the horizontal line test
  • 00:07:58
    the flip side of that is is if a
  • 00:08:00
    function does not pass the horizontal
  • 00:08:01
    line test then it should not have an
  • 00:08:04
    inverse now you may be thinking to
  • 00:08:06
    yourselves now hang on the graphs of
  • 00:08:08
    trig functions should not pass the
  • 00:08:10
    horizontal line test trig functions are
  • 00:08:13
    periodic they repeat themselves over and
  • 00:08:15
    over and over again so if I draw a
  • 00:08:16
    horizontal line I should intersect more
  • 00:08:20
    than one point on any trig function and
  • 00:08:22
    you're right this is 100% true but
  • 00:08:25
    luckily there is a way to get around
  • 00:08:27
    that and the way to get around that
  • 00:08:29
    appears on the next
  • 00:08:31
    page now before we get started I will
  • 00:08:34
    admit this is a very dense page of
  • 00:08:37
    information you can see that from just
  • 00:08:39
    this little excerpt that's on the screen
  • 00:08:41
    right now and as I mentioned we are
  • 00:08:43
    covering six functions six inverse
  • 00:08:46
    functions that we need to be worried
  • 00:08:48
    about but there's a couple things I
  • 00:08:50
    would say to that first of all any time
  • 00:08:53
    that you feel like you need a break
  • 00:08:55
    pause the video and take a break walk
  • 00:08:57
    away for a few minutes stretch your legs
  • 00:08:58
    go to the bathroom have a snack come
  • 00:09:00
    back when you're ready it'll still be
  • 00:09:02
    here the second thing I'll will say is
  • 00:09:04
    although I am going to pay some
  • 00:09:05
    attention to sign and there are a couple
  • 00:09:08
    of occasions where I may have to draw a
  • 00:09:10
    graph potentially I am going to for the
  • 00:09:12
    most part introduce the critical
  • 00:09:15
    information for sign and from that point
  • 00:09:17
    onward because as you can already see
  • 00:09:20
    this information kind of repeats as you
  • 00:09:22
    go through I'm going to indicate where
  • 00:09:25
    things are a little bit different or
  • 00:09:26
    explain why they are different so it
  • 00:09:29
    won't quite as dense as it looks here
  • 00:09:31
    this is mostly for completeness of notes
  • 00:09:34
    so saying let's look at sign now as we
  • 00:09:37
    said sign does not pass the horizontal
  • 00:09:39
    line test normally however if I restrict
  • 00:09:42
    sign to a subdomain meaning instead of
  • 00:09:45
    working with all of the possible inputs
  • 00:09:47
    I only work with the inputs between pi/
  • 00:09:50
    2 and Pi / 2 which for those of you who
  • 00:09:54
    are visually minded if I draw the unit
  • 00:09:57
    circle means I am talking about only
  • 00:10:00
    those
  • 00:10:01
    angles in the right hand half of the
  • 00:10:04
    unit circle since we know that < /2 is
  • 00:10:07
    Down Below on the Y AIS and Pi / 2 is up
  • 00:10:11
    above on the Y
  • 00:10:13
    AIS then we have a one: one function
  • 00:10:17
    once we have a one: one function for
  • 00:10:19
    sign on the subdomain we can Define arc
  • 00:10:22
    sign an inverse the inverse then has to
  • 00:10:25
    satisfy a couple of equations which
  • 00:10:27
    shouldn't be surprising after our
  • 00:10:28
    previous p page and they are s of sin
  • 00:10:31
    inverse of a number a should just give
  • 00:10:33
    us a back again and sin inverse of s of
  • 00:10:36
    an angle Theta should give us the angle
  • 00:10:38
    Theta back again but there are some
  • 00:10:41
    caveats a really here is the output of
  • 00:10:45
    sign it is one of the outputs that s
  • 00:10:48
    gives us because remember when we're
  • 00:10:50
    working with a function and an inverse
  • 00:10:52
    we're going back and forth between the
  • 00:10:54
    inputs and the outputs that means that a
  • 00:10:57
    has to fall into the interval from -1 to
  • 00:11:00
    1 and that is also the domain for arc
  • 00:11:03
    sign or inverse sign however you prefer
  • 00:11:05
    to say it Theta is going to have to be
  • 00:11:09
    between piun over 2 and pi/ 2 because
  • 00:11:11
    that's the Restriction we had to make in
  • 00:11:13
    order to get this Ark sign to begin with
  • 00:11:16
    and Pi / 2 to Pi / 2 is the outputs for
  • 00:11:20
    our Ark sign meaning it is the
  • 00:11:23
    range similarly if we move on to cosine
  • 00:11:27
    by restricting cosine to 0 to Pi which
  • 00:11:31
    in our picture corresponds to the upper
  • 00:11:34
    half of the unit circle up
  • 00:11:37
    top it turns out that we can make Arc
  • 00:11:40
    cosine the inverse cosine and it
  • 00:11:43
    satisfies similar equations to what we
  • 00:11:45
    saw up above the big difference being
  • 00:11:48
    that while a still needs to be between 1
  • 00:11:51
    and 1 so once again the domain for AR
  • 00:11:53
    cosine is -1 to 1 the range now is from
  • 00:11:56
    0 to Pi
  • 00:12:00
    moving to tangent I think it's helpful
  • 00:12:02
    to see a a basic sketch of what tangent
  • 00:12:05
    looks
  • 00:12:07
    like so I have sketched one cycle of
  • 00:12:10
    behavior for tangent you'll notice I've
  • 00:12:12
    included its vertical ASM tootes which
  • 00:12:15
    we know for probably the easiest to work
  • 00:12:18
    with full cycle of tangent happen to be
  • 00:12:21
    at piun / 2 and Pi
  • 00:12:24
    over2 this explains why when we move to
  • 00:12:27
    the subdomain for tangent the one we
  • 00:12:29
    have to restrict it to it is from piun /
  • 00:12:32
    2 to piun / 2 kind of like with s but we
  • 00:12:35
    have to exclude the end points because
  • 00:12:38
    those end points are ASM tootes so we
  • 00:12:40
    can't include them once again once we
  • 00:12:43
    make the Restriction we can define arct
  • 00:12:45
    tangent the inverse function and it
  • 00:12:47
    satisfies the equations but
  • 00:12:49
    now the number a can be any real number
  • 00:12:52
    it can be anything from negative to
  • 00:12:54
    positive Infinity specifically because
  • 00:12:57
    tangent goes from negative to posit of
  • 00:12:59
    infinity and Theta must be in the
  • 00:13:02
    restricted domain from piun 2 to piun /
  • 00:13:07
    2 these first three are all pretty
  • 00:13:10
    straightforward the next couple are
  • 00:13:13
    cosecant and secant now in my classes we
  • 00:13:17
    haven't talked about cosecant and secant
  • 00:13:19
    except in passing so I think the most
  • 00:13:21
    useful thing I can do to begin with is
  • 00:13:23
    just provide you with a basic sketch of
  • 00:13:25
    how I get their pictures
  • 00:13:29
    to do this I begin with graphs of s and
  • 00:13:34
    cosine I've only drawn a little bit of
  • 00:13:36
    them because I'm only going to work with
  • 00:13:38
    a relatively small amount of them and I
  • 00:13:40
    make an observation cosecant is the
  • 00:13:42
    reciprocal of s as long as s is not
  • 00:13:45
    equal to zero that means whenever sign
  • 00:13:48
    is not equal to zero say for example
  • 00:13:50
    when s is equal to 1 cosecant is found
  • 00:13:53
    by taking one and dividing it by the
  • 00:13:55
    value of s so when s is equal to 1
  • 00:13:58
    cosecant is is also equal to 1 it is 1 /
  • 00:14:01
    1 when s is equal to -1 cosecant is also
  • 00:14:05
    equal to 1 because you will take one and
  • 00:14:07
    divide it by the value of s 1
  • 00:14:10
    over1 anytime that s is zero well then
  • 00:14:13
    we're going to have to put in a vertical
  • 00:14:15
    ASM
  • 00:14:18
    toote from here we draw this strange
  • 00:14:21
    looking function wherever it is above
  • 00:14:24
    the xais it cups
  • 00:14:27
    upward whenever it is below the xais it
  • 00:14:31
    cups downward like
  • 00:14:35
    so similarly if we do this for
  • 00:14:40
    secant we can label the points where
  • 00:14:43
    secant is equal to one and negative 1 we
  • 00:14:46
    can label where the vertical ASM toote
  • 00:14:48
    should be and once again we get these
  • 00:14:50
    funny little cupping shapes above the
  • 00:14:52
    x-axis they go up below the xaxis they
  • 00:14:56
    go down
  • 00:15:01
    and it's because of these funny shapes
  • 00:15:02
    that we wind up getting the restrictions
  • 00:15:04
    that we do so you'll notice we have to
  • 00:15:06
    restrict cosecant to the subdomain from
  • 00:15:09
    piun / 2 to Pi / 2 which is a lot like s
  • 00:15:12
    except we can't include zero because
  • 00:15:15
    zero is an ASM toote in our picture
  • 00:15:18
    otherwise it should be the same
  • 00:15:19
    subdomain once we do that we get the
  • 00:15:22
    same basic equations like what we would
  • 00:15:24
    expect the difference is first of all we
  • 00:15:27
    have to restrict Theta to that weird
  • 00:15:30
    little uh subdomain from piun 2 to Pi /
  • 00:15:33
    2 but don't include zero second of all
  • 00:15:36
    we have to say that a is from negative
  • 00:15:38
    Infinity to - 1 or it is from 1 to
  • 00:15:41
    infinity and that comes directly from
  • 00:15:43
    this cupping Behavior you'll notice that
  • 00:15:45
    we never find secant between negative 1
  • 00:15:48
    and 1 it is either negative 1 and Below
  • 00:15:51
    or positive 1 and above and that
  • 00:15:53
    explains why a is in this funny looking
  • 00:15:55
    interval here once again we are looking
  • 00:15:57
    at the Domain and the range for our
  • 00:16:00
    inverse function similarly when working
  • 00:16:03
    with secant we find out that we really
  • 00:16:05
    want to mimic the behavior of cosine
  • 00:16:08
    it's just that if we go from 0 to Pi
  • 00:16:10
    we're going to wind up with this
  • 00:16:11
    vertical ASM toote in the middle so we
  • 00:16:13
    go from 0 to Pi and remove the vertical
  • 00:16:16
    asmp toote at Pi / 2 after that we wind
  • 00:16:19
    up with the standard looking equations
  • 00:16:22
    we get the domain of arc secant which is
  • 00:16:24
    again from negative Infinity to 1 and
  • 00:16:27
    from 1 to infinity and and we restrict
  • 00:16:30
    Theta from 0 to Pi excluding the ASM
  • 00:16:33
    toote Pi /
  • 00:16:36
    2 after these two functions I would say
  • 00:16:39
    that coent is relatively quite easy once
  • 00:16:42
    again let me go ahead and sketch
  • 00:16:47
    coent coent has a very similar shape to
  • 00:16:50
    tangent the difference being tangent
  • 00:16:52
    increases over its cycle whereas Cent
  • 00:16:55
    clearly decreases Cent has an AS toote
  • 00:16:59
    at zero and another ASM toote at pi and
  • 00:17:02
    this explains the subdomain we're going
  • 00:17:04
    through a full cycle of behavior from 0
  • 00:17:07
    to Pi but not including the asmp tootes
  • 00:17:10
    once we do that restriction we are able
  • 00:17:12
    to put down these predictable equations
  • 00:17:16
    once again cotangent goes down to
  • 00:17:18
    negative Infinity up to positive
  • 00:17:19
    Infinity so a must belong to negative
  • 00:17:22
    Infinity to
  • 00:17:24
    infinity and Theta must belong to 0 to
  • 00:17:27
    Pi and these are once again let me go
  • 00:17:29
    ahead and state it over and over again
  • 00:17:31
    these are the domain and the range of
  • 00:17:33
    Arc cotangent we're going to need these
  • 00:17:36
    domains and ranges and this is why I've
  • 00:17:38
    labeled them this way now at this point
  • 00:17:41
    we are almost ready to do examples which
  • 00:17:44
    is of course the thing we want to do
  • 00:17:46
    most but before we do that we have one
  • 00:17:48
    more page of information we need to go
  • 00:17:50
    over and it's just to prepare you for
  • 00:17:52
    the
  • 00:17:55
    examples I have some major equations for
  • 00:17:58
    each of the six inverse or Arc trig
  • 00:18:01
    functions now at this point I would say
  • 00:18:03
    just pause the video and write them all
  • 00:18:05
    down so you can refer to them later I
  • 00:18:07
    will also say to you these are not the
  • 00:18:09
    only ones there are some others for
  • 00:18:12
    example for USF students in our textbook
  • 00:18:14
    you will quickly find that there are
  • 00:18:15
    more equations you'll notice that I have
  • 00:18:18
    green equations that talk about what Ark
  • 00:18:21
    sign for example does with negative a
  • 00:18:23
    but I haven't done that for Arc cosecant
  • 00:18:26
    there is an equation for Arc cosecant
  • 00:18:28
    that looks kind of similar however I
  • 00:18:31
    have not included it the reason for that
  • 00:18:33
    is actually pretty simple when I'm
  • 00:18:36
    working with these inverse trig
  • 00:18:37
    functions these are the ones that I use
  • 00:18:40
    predominantly the ones in
  • 00:18:42
    Black sometimes I will use the green
  • 00:18:44
    ones mostly the green ones will be taken
  • 00:18:46
    care of by sketches as I'll show you
  • 00:18:48
    when we go through the examples but I
  • 00:18:50
    include them both because I can match up
  • 00:18:54
    these equations with the sketches that I
  • 00:18:56
    draw later on when we get to the
  • 00:18:57
    examples and also because sometimes the
  • 00:19:00
    homework software will ask you to use
  • 00:19:02
    one of these equations so it's important
  • 00:19:04
    to have them in front of you and to
  • 00:19:06
    think about them and know what they
  • 00:19:09
    are now I don't want to beat a dead
  • 00:19:11
    horse with these uh equations that I've
  • 00:19:14
    written up here I am 100% confident that
  • 00:19:16
    all of you can read the equations that
  • 00:19:18
    I've written so what I'm going to do is
  • 00:19:21
    I'm simply going to summarize the
  • 00:19:22
    information that you see here once again
  • 00:19:25
    kind of like we talked about on the
  • 00:19:26
    first page whenever I have a function
  • 00:19:29
    that has an inverse even if I had to do
  • 00:19:30
    a little bit of restricting in order to
  • 00:19:32
    get to this point I can write
  • 00:19:35
    information about the functions two
  • 00:19:37
    different ways I can either say the
  • 00:19:39
    function of an input is equal to an
  • 00:19:41
    output or I can say the inverse function
  • 00:19:44
    of the output is equal to the input the
  • 00:19:46
    Only Rule here is a must be in the
  • 00:19:53
    domain so if you go back to the previous
  • 00:19:56
    page and look through that whole list of
  • 00:19:58
    function and definitions there whatever
  • 00:20:00
    the domain is for your particular
  • 00:20:02
    inverse trig function a must be in that
  • 00:20:05
    domain
  • 00:20:06
    andeta must be in the
  • 00:20:10
    range if that doesn't happen these
  • 00:20:13
    equations that I've written here don't
  • 00:20:14
    make any sense anymore and they can't be
  • 00:20:16
    used so you have to keep that in mind
  • 00:20:18
    when you're working on these things as
  • 00:20:21
    long as we have that information here I
  • 00:20:23
    think that we are ready to do our very
  • 00:20:24
    first
  • 00:20:27
    example now for those of you students uh
  • 00:20:29
    at USF you will notice that I have done
  • 00:20:32
    probably a whole bunch of problems at
  • 00:20:34
    this point that look like the one I've
  • 00:20:35
    done here calculating arc sign for
  • 00:20:38
    various uh values here I still want to
  • 00:20:43
    include it in our notes because I feel
  • 00:20:44
    like it is important to do these
  • 00:20:46
    examples over and over again until it
  • 00:20:48
    clicks also we've not used these
  • 00:20:50
    particular values so this is good
  • 00:20:53
    practice I'm going to go ahead and start
  • 00:20:55
    with the Positive value for a so I am
  • 00:20:58
    going to start
  • 00:20:59
    with sin inverse or S of < tk3 /
  • 00:21:06
    2 and the very first thing I'm going to
  • 00:21:08
    do we've mentioned this in class is I'm
  • 00:21:11
    going to double check that this value
  • 00:21:13
    actually belongs to the to the domain
  • 00:21:16
    now I happen to remember the domain for
  • 00:21:18
    AR sign go back if you don't but arc
  • 00:21:22
    sign has a domain of -1 to 1 and a
  • 00:21:25
    little bit of calculation if you have to
  • 00:21:28
    put it into to a calculator will show
  • 00:21:30
    you that < tk3 over2 belongs to that
  • 00:21:33
    domain so it is okay to talk about AR
  • 00:21:36
    sign of < tk3 over2 and we should get an
  • 00:21:40
    answer for now I'm going to put Theta in
  • 00:21:44
    the answers Place Theta is just a dummy
  • 00:21:46
    variable meaning not that I'm a dummy or
  • 00:21:48
    at least I hope I'm not but rather that
  • 00:21:50
    I'm filling in the space because now
  • 00:21:53
    that I've written the equation AR sign
  • 00:21:55
    of a number equals Theta which is kind
  • 00:21:57
    of like the right hand side of this
  • 00:21:59
    major equation up here I can now write
  • 00:22:02
    the other
  • 00:22:05
    perspective the other perspective would
  • 00:22:07
    be if I take s of theta I should be
  • 00:22:10
    getting < tk3
  • 00:22:15
    over2 so what we're looking for now just
  • 00:22:17
    to make sure you remember is an angle
  • 00:22:20
    Theta the angle Theta has to be in the
  • 00:22:22
    range of arc sign again I'll let you go
  • 00:22:25
    back and check but the range for arc
  • 00:22:27
    sign should be the right hand half of
  • 00:22:30
    the unit circle from negative pi over2
  • 00:22:32
    to positive Pi / 2 so are there any
  • 00:22:35
    angles between negative pi over2 and Pi
  • 00:22:37
    / 2 that give us < tk3 over2 when I plug
  • 00:22:40
    them into sign absolutely if you
  • 00:22:44
    recall s of pi over 3 which in other
  • 00:22:48
    words means s of
  • 00:22:50
    60° is equal to < tk3
  • 00:22:54
    over2 in fact if you look through the
  • 00:22:57
    right hand half of the unit circle
  • 00:22:59
    meaning quadrant 1 and Quadrant 4 you're
  • 00:23:02
    going to quickly find out that that
  • 00:23:03
    doesn't happen anywhere else pi over 3
  • 00:23:06
    is the only angle for which s gives us <
  • 00:23:09
    tk3 over2 so with this in mind there's
  • 00:23:12
    only one possible answer for Theta as
  • 00:23:15
    long as Theta is following the rules as
  • 00:23:17
    long as Theta belongs to the range of my
  • 00:23:19
    arc sign Theta must be pi over
  • 00:23:25
    3 Let's scoot down a little bit I'm
  • 00:23:27
    going to change colors to Blue and we're
  • 00:23:30
    going to do this again but now we're
  • 00:23:32
    going to do it for the negative value so
  • 00:23:34
    we will say we're taking arine of < tk3
  • 00:23:43
    over2 once again < tk3 over2 is in the
  • 00:23:47
    domain of Ark sign it is between 1 and 1
  • 00:23:50
    so I'm going to put a dummy variable
  • 00:23:53
    once I have written an equation I'm
  • 00:23:55
    going to write the other perspective on
  • 00:23:57
    that equation that if I take sign of
  • 00:24:00
    whatever this angle is I must be getting
  • 00:24:02
    out of
  • 00:24:03
    it < tk3
  • 00:24:06
    over2 and I'm going to stop and think
  • 00:24:08
    about this one for this one I really
  • 00:24:10
    need to have a diagram so I'm just going
  • 00:24:12
    to really quickly sketch
  • 00:24:14
    one now last time when we were drawing
  • 00:24:18
    things we wound up with a an answer an
  • 00:24:21
    angle that was in the first quadrant you
  • 00:24:24
    should go and check this for yourself
  • 00:24:26
    but Pi 3 is definitely in quadrant 1 and
  • 00:24:30
    that makes sense because in quadrant 1
  • 00:24:32
    every single one of the trig functions
  • 00:24:33
    is positive however this time around
  • 00:24:36
    we're getting negative
  • 00:24:38
    answers since we are stuck with the
  • 00:24:40
    right hand side of the unit circle
  • 00:24:42
    between Pi / 2 and positive Pi / 2
  • 00:24:47
    there's only one place where this angle
  • 00:24:49
    can live if it is going to make any
  • 00:24:51
    sense and that angle is here in Quadrant
  • 00:24:55
    4 specifically we must get to it by
  • 00:24:58
    rotating the wrong way into Quadrant
  • 00:25:02
    4 H now what angle could do
  • 00:25:07
    that oh now wait a second if I rotate to
  • 00:25:11
    an angle like this in Quadrant
  • 00:25:13
    4 notice s of piun / 3 is postive < tk3
  • 00:25:18
    /
  • 00:25:19
    2 and what that's implying to us is
  • 00:25:22
    because of the way s works because s is
  • 00:25:25
    an odd function and it goes positive
  • 00:25:28
    when I go go this way but negative when
  • 00:25:29
    I go this way if I take S ofk
  • 00:25:36
    3 I must get < tk3
  • 00:25:43
    over2 because of that and because kind
  • 00:25:45
    of like last time if you start checking
  • 00:25:47
    around you find out that there is no
  • 00:25:50
    other angle between piun / 2 and pi over
  • 00:25:52
    2 that gives me this result it must be
  • 00:25:55
    that the answer for this particular
  • 00:25:57
    equation is that Theta must beunk over 3
  • 00:26:01
    or in other words Ark sign of < tk3 /2
  • 00:26:05
    is equal to pi over
  • 00:26:08
    3 now I want you to take a look at these
  • 00:26:10
    answers last time when I took Ark sign
  • 00:26:13
    of ro3 over2 I got pi over 3 this time
  • 00:26:17
    when I took Ark sign of NE < tk3 over2 I
  • 00:26:21
    got netive piun
  • 00:26:23
    over3 that is exactly what this green
  • 00:26:25
    equation says here if I take Ark of the
  • 00:26:29
    negative of a then the result must be
  • 00:26:32
    the negative of whatever AR sign of a
  • 00:26:34
    gave
  • 00:26:36
    me in other words we've just recreated
  • 00:26:39
    the green equation for Ark sign we've
  • 00:26:41
    just done it using diagrams and using
  • 00:26:43
    our work here that should not be too
  • 00:26:47
    surprising okay let's go ahead and do
  • 00:26:49
    another
  • 00:26:52
    example in fact we're going to go ahead
  • 00:26:55
    and do the same exercise again but this
  • 00:26:57
    time we're going to work with Arc cosine
  • 00:26:59
    instead of Co instead of AR
  • 00:27:02
    s now I'm going to go ahead and show the
  • 00:27:05
    steps again a little bit faster but
  • 00:27:07
    after we do this example I'm going to
  • 00:27:09
    speed things up a bit so be prepared
  • 00:27:12
    make sure you're taking notes
  • 00:27:14
    here once again I'm taking Arc cosine
  • 00:27:18
    I'm taking it of let's say the positive
  • 00:27:21
    first I'm checking to make sure that Roo
  • 00:27:23
    tk3 over2 belongs to The Domain it does
  • 00:27:26
    we happen to know the domain for Arc
  • 00:27:28
    cosine and AR sign are identical and
  • 00:27:30
    we're setting up our dummy variable so
  • 00:27:32
    that we can rewrite our information from
  • 00:27:35
    the other perspective that is for this
  • 00:27:39
    first equation to be true it must also
  • 00:27:41
    be true that cosine of the angle Theta
  • 00:27:43
    whatever Theta is must be giving us <
  • 00:27:45
    tk3
  • 00:27:47
    over2 this is going to give us a
  • 00:27:49
    different answer than last time however
  • 00:27:52
    if you start looking through the range
  • 00:27:55
    for Arc cosine which I will remind you
  • 00:27:57
    is the upper half of the unit circle
  • 00:28:00
    first of all because we have a positive
  • 00:28:02
    answer here we must be talking about an
  • 00:28:04
    angle in quadrant one because quadrant 2
  • 00:28:07
    is where cosine is negative or I should
  • 00:28:09
    say one of the places where cosine is
  • 00:28:11
    negative second of all I happen to know
  • 00:28:15
    from experience that 30° which is to say
  • 00:28:20
    pi/
  • 00:28:21
    6 is an angle for which we get < tk3
  • 00:28:24
    over2 when we plug it into cosine and in
  • 00:28:28
    fact it is the only angle in the upper
  • 00:28:31
    half of the circle that does that so it
  • 00:28:33
    must follow that Theta is equal to pi/ 6
  • 00:28:40
    30° giving myself a little bit of room
  • 00:28:44
    if I do the same thing again but this
  • 00:28:46
    time with my AR cosine of negative < tk3
  • 00:28:52
    over2 I check my domain considerations
  • 00:28:55
    they're good I write an equation with a
  • 00:28:58
    dummy variable so I can write my other
  • 00:29:06
    perspective I note looking at this
  • 00:29:09
    second equation that cosine of theta is
  • 00:29:11
    giving me a negative and since we are
  • 00:29:13
    restricted to the upper half of the unit
  • 00:29:15
    circle which is quadrants 1 and
  • 00:29:17
    two this implies to me that Theta must
  • 00:29:20
    belong to quadrant 2 so the question
  • 00:29:22
    then is what is an angle in quadrant 2
  • 00:29:25
    that does what we
  • 00:29:26
    want well last time when we were working
  • 00:29:29
    with Ark sign I didn't say the word
  • 00:29:34
    reference angle but I'm going to say it
  • 00:29:36
    now because here's what I
  • 00:29:42
    notice if I draw an angle in quadrant
  • 00:29:48
    2 this will be my Theta right
  • 00:29:52
    here I will get negative as I want it if
  • 00:29:56
    I want to make sure that Co of theta
  • 00:29:58
    gives
  • 00:29:59
    mek3 /2 one thing that I could appeal to
  • 00:30:04
    is reference
  • 00:30:06
    angles I already know that cine of piun
  • 00:30:09
    / 6 gives me positive < tk3 over2 so if
  • 00:30:12
    I ensure that the reference angle for
  • 00:30:15
    Theta is pi/ 6 then cosine of theta
  • 00:30:20
    should basically be the same as pi over
  • 00:30:22
    6 except because it is in quadrant 2 it
  • 00:30:27
    will be negative so here's a question
  • 00:30:29
    what angle in quadrant 2 has a reference
  • 00:30:32
    angle of Pi / 6 well there's actually a
  • 00:30:36
    pretty easy way to set that up and
  • 00:30:37
    figure that
  • 00:30:40
    out since we are drawing an
  • 00:30:44
    angle Theta here in quadrant 2 and since
  • 00:30:47
    the reference angle that we've indicated
  • 00:30:49
    is between Theta and Pi we could always
  • 00:30:52
    set up an equation that looks like this
  • 00:30:54
    when I take pi and I subtract Theta I
  • 00:30:58
    should get Pi / 6 the reference angle if
  • 00:31:01
    we solve this for Theta we wind up
  • 00:31:05
    finding that Theta must be equal to piun
  • 00:31:09
    - < / 6 and if you run the calculations
  • 00:31:12
    on that that means Theta must be equal
  • 00:31:15
    to 5 pi/ 6 so coming back up to my
  • 00:31:20
    work I write down well cine of 5 piun /
  • 00:31:26
    6 is equal to < tk3 / 2 as desired so we
  • 00:31:32
    must find that Theta is equal to 5 piun
  • 00:31:37
    over 6 and that completes the
  • 00:31:39
    exercise now last time when we got to
  • 00:31:42
    this point we compared our answers last
  • 00:31:44
    time we found R cosine ofun 33/2 was
  • 00:31:47
    piun / 6 this time we found R cosine of
  • 00:31:51
    negative < tk3 over2 was 5 pi over
  • 00:31:54
    6 really the comparison though is
  • 00:31:57
    between the previous answer in red and
  • 00:31:59
    what we wrote down here below our sketch
  • 00:32:02
    you notice that when we figured out what
  • 00:32:04
    Theta what our answer would be we had to
  • 00:32:07
    take pi and subtract pi/ 6 or in other
  • 00:32:10
    words we had to take pi and subtract our
  • 00:32:12
    old answer well if we go back up again
  • 00:32:16
    you'll notice that the companion
  • 00:32:17
    equation in green for Arc cosine is if I
  • 00:32:20
    want to do AR cosign of the negative of
  • 00:32:23
    some number a I can find it by taking pi
  • 00:32:26
    and subtracting R Co of a for us that
  • 00:32:29
    means R cosine of3 /2 is the same thing
  • 00:32:33
    as Pi minus r cosine of POS < tk3 over2
  • 00:32:37
    and you'll notice that we've recovered
  • 00:32:39
    this property as
  • 00:32:40
    well it's one of the reasons why we
  • 00:32:42
    really don't need those properties even
  • 00:32:44
    though it's good to talk about them we
  • 00:32:46
    can reproduce them even without having
  • 00:32:48
    to have the formulas in front of
  • 00:32:52
    us let's do an example for arc tangent
  • 00:32:56
    you'll notice in this one I've only
  • 00:32:57
    given one value that I want to check I
  • 00:33:00
    want to make sure we speed things up I
  • 00:33:02
    think by now you should be getting the
  • 00:33:04
    gist of how we do these things so in
  • 00:33:06
    fact as I promised I'm going to do this
  • 00:33:08
    kind of quickly pause the video and try
  • 00:33:10
    to see if you can fill in the gaps that
  • 00:33:12
    I'm about to fill in and see where you
  • 00:33:17
    are okay basic stuff out of the way we
  • 00:33:20
    know that the domain for arc tangent is
  • 00:33:23
    all real numbers so there's really
  • 00:33:25
    nothing to check there if we write the
  • 00:33:27
    dummy equation and then write the other
  • 00:33:29
    perspective we find that tangent of
  • 00:33:31
    should be equal
  • 00:33:32
    tok3 and once again we're dealing with a
  • 00:33:36
    situation where we must have an angle in
  • 00:33:39
    a different quadrant than quadrant one
  • 00:33:41
    we know that tangent is going to be
  • 00:33:44
    restricted to an arc tangent has a range
  • 00:33:46
    of negative piun over 2 to positive piun
  • 00:33:49
    / 2 and clearly if we were in quadrant
  • 00:33:51
    one we would have a positive result
  • 00:33:54
    since we're getting a negative result
  • 00:33:56
    this must mean that we are going to be
  • 00:33:58
    dealing with something in Quadrant 4
  • 00:34:00
    going in the negative
  • 00:34:03
    Direction sketching out a picture as I'm
  • 00:34:06
    want to do when I need some
  • 00:34:09
    ideas I must be talking about an angle
  • 00:34:13
    somewhere down here like so going this
  • 00:34:21
    way now I have to think through some of
  • 00:34:24
    my reference angles here I need to think
  • 00:34:26
    through the basic angles like the
  • 00:34:28
    quadrantal angles those would be big
  • 00:34:30
    ones for me and then the angles pi/ 6 30
  • 00:34:33
    uh 30° pi over 4 45° and pi over 3 which
  • 00:34:38
    is
  • 00:34:39
    60° and as I think through those I have
  • 00:34:42
    a
  • 00:34:43
    recollection I
  • 00:34:47
    recall that the angle pi over 3 in other
  • 00:34:51
    words 60° has an x coordinate of 12 and
  • 00:34:55
    a y coordinate of < tk3 two on the unit
  • 00:35:00
    circle since tangent is given by the y
  • 00:35:04
    coordinate divided by the x
  • 00:35:07
    coordinate this tells me with a little
  • 00:35:09
    bit of simple calculation that tangent
  • 00:35:11
    of pi over 3 must be equal or excuse me
  • 00:35:15
    tangent of pi over 3 must be equal to
  • 00:35:17
    positive < tk3 which is almost what we
  • 00:35:21
    want here's another observation if you
  • 00:35:24
    notice if I were to make this angle in
  • 00:35:27
    red the one I'm trying to figure
  • 00:35:30
    outk
  • 00:35:32
    3 notice that that would mean that these
  • 00:35:35
    two angles the pi over 3 and the
  • 00:35:37
    negative pi over 3 should be direct
  • 00:35:39
    reflections of one another across the x
  • 00:35:43
    axis whenever that happens we know that
  • 00:35:47
    the y coordinate would change so we
  • 00:35:49
    would have still2 for the x coordinate
  • 00:35:52
    but now a NE < tk3 over2 for the
  • 00:35:56
    y-coordinate
  • 00:35:58
    since the y-coordinate is negative and
  • 00:36:01
    the x coordinate is positive this
  • 00:36:03
    implies that tangent of pi over 3 will
  • 00:36:05
    be a negative number but otherwise it
  • 00:36:07
    should give us pretty much the same
  • 00:36:09
    answer so in fact what we figured out
  • 00:36:11
    from this diagram is tangent of piun
  • 00:36:14
    over 3 gives us the value we
  • 00:36:19
    want listen to that my dog is
  • 00:36:21
    celebrating in the background you go
  • 00:36:25
    girl and that must mean that Thea is
  • 00:36:28
    equal to negative or excuse me
  • 00:36:32
    a no I was right the first time man I
  • 00:36:35
    got distracted by the dog what can I say
  • 00:36:38
    dogs distract people Theta must be equal
  • 00:36:41
    to piun over 3 and there you go we've
  • 00:36:44
    got our
  • 00:36:46
    example now that we have the basic
  • 00:36:49
    examples for Ark sign through arc
  • 00:36:52
    tangent I want to do a couple of
  • 00:36:54
    problems for the other weird functions
  • 00:37:01
    part of this is because they are a
  • 00:37:02
    little bit weird but also to show you
  • 00:37:05
    the rules that I use for evaluating
  • 00:37:09
    them oh she's singing now that's very
  • 00:37:11
    sweet all right sing us into the next
  • 00:37:14
    problem oh sweet
  • 00:37:16
    doggo oky doie we have our next example
  • 00:37:21
    calculate Arc secant of two now I'm
  • 00:37:25
    going to leave it up to you to double
  • 00:37:27
    double check that two falls into the
  • 00:37:30
    domain for arant you've got to get used
  • 00:37:32
    to doing this for yourself so I'm going
  • 00:37:33
    to go ahead and let you do that part I'm
  • 00:37:36
    also going to use this time to remind
  • 00:37:40
    you once again just like with all the
  • 00:37:42
    previous examples we must have another
  • 00:37:45
    perspective that we can
  • 00:37:48
    write and I guess more accurately what I
  • 00:37:51
    really want to do is to take this to
  • 00:37:52
    remind you you should be getting used to
  • 00:37:54
    writing these yourself I've taken the
  • 00:37:56
    liberty of doing it here make sure that
  • 00:37:58
    you're writing these yourself now what I
  • 00:38:01
    need to do from here is a little bit
  • 00:38:03
    different from the last few times I
  • 00:38:05
    really don't want to have to work with
  • 00:38:06
    secant because secant is a bit weird
  • 00:38:08
    being a reciprocal function so instead
  • 00:38:11
    I'm going to make an
  • 00:38:13
    observation if secant of theta is equal
  • 00:38:15
    to 2 because I know that cosine and
  • 00:38:18
    secant are reciprocals of one another it
  • 00:38:21
    suggests that cosine of theta must be
  • 00:38:24
    equal to 12
  • 00:38:28
    if we can work with cosine of theta is
  • 00:38:29
    equal to 12 well that's a much easier
  • 00:38:32
    situation to work with in fact if you
  • 00:38:34
    know your basic reference angles pretty
  • 00:38:37
    well then you should know which angle to
  • 00:38:39
    look
  • 00:38:41
    for reaching through my memory here I
  • 00:38:45
    recall that
  • 00:38:49
    cosine of pi over
  • 00:38:53
    3 is equal to 12
  • 00:38:58
    I am that good I think it should be
  • 00:39:00
    recorded on video but on top of that you
  • 00:39:03
    can actually see up above I drew the
  • 00:39:05
    angle pi over 3 and look the x
  • 00:39:07
    coordinate is 1/ 12 and cosine is the x
  • 00:39:09
    coordinate points on the unit circle so
  • 00:39:12
    with that information in toe I know that
  • 00:39:15
    cosine is of Pi 3 is what I want for
  • 00:39:18
    this line since pi over 3 makes cosine
  • 00:39:21
    equal to 12 it also makes secant equal
  • 00:39:23
    to two and it's the only angle that does
  • 00:39:26
    that
  • 00:39:28
    in well in the range that we're looking
  • 00:39:32
    at which we should go ahead and write
  • 00:39:35
    down because I realized I hadn't written
  • 00:39:37
    that
  • 00:39:39
    down the range that we're looking at
  • 00:39:42
    should be the upper half of the unit
  • 00:39:44
    circle except for except for the line
  • 00:39:49
    where Theta is equal to Pi /
  • 00:39:51
    2 so since this is the only angle that's
  • 00:39:55
    going to give us a satisfactory answer
  • 00:39:57
    it follows at once that our answer to
  • 00:39:59
    the problem is that we must be getting
  • 00:40:02
    pi over 3 see inverse of 2 must be equal
  • 00:40:05
    to pi over
  • 00:40:07
    3 we're going to do this again let's go
  • 00:40:09
    ahead and do it for
  • 00:40:13
    cosecant all right if we're doing this
  • 00:40:15
    problem for cosecant we have to do a lot
  • 00:40:17
    of the same setup we are going to have
  • 00:40:19
    to start out and verify that two does
  • 00:40:22
    indeed belong to the domain of cosecant
  • 00:40:24
    AR cosecant I should say let's be
  • 00:40:26
    precise here here we are also going to
  • 00:40:28
    have to write the other perspective
  • 00:40:30
    equation and we are going to have to
  • 00:40:32
    write out another equation this time
  • 00:40:35
    because cosecant is the reciprocal of s
  • 00:40:39
    it won't be cosine of theta that we want
  • 00:40:41
    to work with but co uh but s of theta so
  • 00:40:44
    I'm going to let you go ahead and write
  • 00:40:46
    that stuff down I'm going to return with
  • 00:40:48
    my answers in just a
  • 00:40:51
    moment we're back so you can see how
  • 00:40:54
    I've written everything down you see
  • 00:40:56
    that we should be finding out that
  • 00:40:58
    whatever our answer is if I plug that
  • 00:41:00
    angle into sign I should be getting 1/2
  • 00:41:03
    and I know that when I'm looking at ARC
  • 00:41:05
    cosecant and its answers the result
  • 00:41:07
    should be somewhere on the right hand
  • 00:41:09
    side of the unit circle excluding the
  • 00:41:12
    line where Theta is equal to zero now
  • 00:41:15
    once again I've chosen this value two so
  • 00:41:18
    I get 1/2 and that's because 1/2 happens
  • 00:41:21
    to correspond with one of our common
  • 00:41:22
    reference angles
  • 00:41:25
    specifically if you do a little bit of
  • 00:41:26
    Cal calculation you'll find out that s
  • 00:41:30
    ofk / 6 s of 30° is equal to
  • 00:41:35
    12 and so the answer we should be
  • 00:41:38
    getting for this problem is pi 6
  • 00:41:43
    predictably this shows you between these
  • 00:41:45
    two examples the basic Theory as it
  • 00:41:47
    applies to Arc cosecant Arc secant and
  • 00:41:51
    even our cotangent although I will leave
  • 00:41:53
    that example for you to try out what you
  • 00:41:56
    really need to do is get to the point
  • 00:41:57
    where you've written the second
  • 00:41:59
    perspective on the same information then
  • 00:42:02
    rewrite the equation using the
  • 00:42:03
    reciprocal function for secant that's
  • 00:42:06
    cosine for cosecant that's s for coent
  • 00:42:10
    because of the uh reciprocal identities
  • 00:42:14
    you're going to have to go and look
  • 00:42:15
    those up in the textbook but because of
  • 00:42:17
    the reciprocal identities for cotangent
  • 00:42:19
    that will be tangent once you write that
  • 00:42:21
    equation you will know what to do from
  • 00:42:24
    there this should be enough examples of
  • 00:42:26
    this type of problem let's move on to
  • 00:42:28
    some more challenging and more
  • 00:42:29
    interesting
  • 00:42:32
    ones scrolling down just a little bit we
  • 00:42:35
    have our next problem we want to
  • 00:42:37
    calculate sin inverse of s of pi over 3
  • 00:42:40
    this is one of the common types of
  • 00:42:41
    problems you're going to run into when
  • 00:42:42
    you're doing your homework problems we
  • 00:42:45
    have to figure out how to evaluate this
  • 00:42:48
    now there is a fast way to do this and
  • 00:42:50
    so I'm going to do the fast way because
  • 00:42:52
    I think it will help you remember some
  • 00:42:54
    important properties but if you remember
  • 00:42:56
    I'll just just put this here as a little
  • 00:42:58
    reminder to go back to that second page
  • 00:43:00
    of notes that we made apparently eons
  • 00:43:04
    ago when I take s inverse of s of an
  • 00:43:10
    angle Theta the result is going to be
  • 00:43:13
    Theta
  • 00:43:16
    if Theta is
  • 00:43:18
    in the restricted domain well I guess I
  • 00:43:22
    should say either in the subdomain for
  • 00:43:24
    sign or in the range of Arc sign which
  • 00:43:27
    in our case means between PK /
  • 00:43:30
    2 and < /
  • 00:43:35
    2 we wrote this down on the previous
  • 00:43:37
    page when we defined these functions and
  • 00:43:39
    it's important here because as I look at
  • 00:43:42
    my problem here I realize pi over 3 is
  • 00:43:45
    in this interval specifically pi over 3
  • 00:43:48
    is a quadrant one angle so it makes
  • 00:43:51
    perfect sense that it should be less
  • 00:43:52
    than Pi / 2 and it's definitely bigger
  • 00:43:54
    thank / 2 with that that said I really
  • 00:43:58
    don't need to do any
  • 00:44:00
    calculation the answer is just pi over
  • 00:44:03
    3 but I would also like to go ahead and
  • 00:44:07
    sketch a really quick picture for you
  • 00:44:08
    because I think it helps to illustrate
  • 00:44:10
    something we're going to need in a
  • 00:44:13
    moment all right I have gone ahead and
  • 00:44:15
    drawn the angle pi over 3 and I've
  • 00:44:18
    included also its terminal point and
  • 00:44:21
    here's why I say that remember that
  • 00:44:24
    What's Happening Here is we are
  • 00:44:26
    basically going through through steps if
  • 00:44:28
    we were to follow our diagram like we
  • 00:44:30
    did on the first page what would happen
  • 00:44:32
    is we would first plug pi over 3 into s
  • 00:44:35
    and it would take us to this value here
  • 00:44:39
    then what we would do is we would want
  • 00:44:41
    to find an angle that gives us this
  • 00:44:43
    value when we plug the angle into sign
  • 00:44:46
    which must be the angle that goes along
  • 00:44:47
    with it here in my drawing now it's kind
  • 00:44:50
    of useless here because we have this
  • 00:44:53
    nice information sin inverse of sin of
  • 00:44:56
    theta equal Theta
  • 00:44:57
    specifically because Pi 3 belongs to the
  • 00:45:01
    uh the range of Ark sign but in just a
  • 00:45:04
    little bit we're going to see problems
  • 00:45:06
    where it's not quite so clear-cut so I
  • 00:45:07
    want to get used to drawing these
  • 00:45:09
    pictures you're going to need them again
  • 00:45:11
    in a
  • 00:45:13
    moment our next example calculate
  • 00:45:16
    tangent of tangent inverse tangent of
  • 00:45:19
    arc tangent of pi over 3 now once again
  • 00:45:22
    there's something important to recall so
  • 00:45:24
    I'll go ahead and put that here
  • 00:45:29
    remember from page two when I take
  • 00:45:32
    tangent of arc tangent of a we get a
  • 00:45:34
    again as long as a belongs to the domain
  • 00:45:39
    of arc tangent it's between negative
  • 00:45:41
    infinity and infinity now in this
  • 00:45:44
    particular case that makes things really
  • 00:45:46
    easy again we really don't have much to
  • 00:45:48
    worry
  • 00:45:49
    about this just says a has to be a real
  • 00:45:52
    number well it's pi over 3 a real number
  • 00:45:54
    you beta so when I come over here and WR
  • 00:45:57
    the answer is pi over 3 the tangent and
  • 00:46:00
    the arct tangent undo each other and I'm
  • 00:46:02
    simply left with whatever a was that's
  • 00:46:04
    all there is to
  • 00:46:06
    it but now is when things are going to
  • 00:46:08
    get a little bit tricky and this is
  • 00:46:10
    where we're going to have to start
  • 00:46:11
    paying attention to things because it's
  • 00:46:13
    not always this
  • 00:46:16
    clean for example check out this problem
  • 00:46:19
    right here cosine of cosine inverse of
  • 00:46:23
    pi/ 3 calculate if it exists you'll
  • 00:46:27
    notice I've put the hint if it exists
  • 00:46:29
    here because I'm not going to be leaving
  • 00:46:30
    you in suspense for very long and the
  • 00:46:32
    answer is it doesn't here's the basic
  • 00:46:35
    reason
  • 00:46:38
    why recalling from page two that cosine
  • 00:46:42
    of Arc coine of a is equal to a if a
  • 00:46:46
    belongs to -1 to 1 we now really have to
  • 00:46:50
    ask ourselves a question we know the pi
  • 00:46:53
    over 3 is a real number but is it in
  • 00:46:55
    this interval well actually a quick
  • 00:46:57
    estimate will show us it is not it's an
  • 00:47:00
    easy one too when looking at the number
  • 00:47:03
    pi over 3 I happen to know that Pi is
  • 00:47:06
    bigger than three it's
  • 00:47:09
    3.1415926 etc etc so what I could always
  • 00:47:13
    do is say Hey listen if I have the
  • 00:47:15
    fraction pi over 3 you know it's a
  • 00:47:17
    smaller fraction three instead of pi
  • 00:47:20
    over three and guess what 3 over 3 is
  • 00:47:24
    equal to 1
  • 00:47:31
    so there is no answer in this
  • 00:47:34
    case we've picked a number that is
  • 00:47:37
    outside of the domain of AR cosine so
  • 00:47:40
    the this does not
  • 00:47:43
    exist there is no
  • 00:47:47
    answer you have to watch out for that
  • 00:47:50
    some of them like for example tangent
  • 00:47:52
    are very easy to deal with and they
  • 00:47:53
    don't have this problem very frequently
  • 00:47:55
    some of them like cosine sign have them
  • 00:47:57
    a lot you have to pay close attention to
  • 00:48:00
    what you're writing down here's another
  • 00:48:02
    interesting problem for you to work
  • 00:48:06
    with this time we want to calculate
  • 00:48:09
    cosine inverse of cosine of 4 pi/ 3 now
  • 00:48:14
    at this point many of you probably
  • 00:48:16
    looked at the problem and said oh
  • 00:48:17
    obviously the answer is 4 pi over
  • 00:48:21
    3 but in fact this is wrong it is not 4
  • 00:48:25
    pi over 3
  • 00:48:27
    the reason why is that the equation that
  • 00:48:29
    you're thinking of the cosine inverse of
  • 00:48:32
    cosine of an angle gives us the angle
  • 00:48:34
    back only works if the angle is in the
  • 00:48:37
    subdomain for cosine in the range of Arc
  • 00:48:41
    cosine which we know is 0 to Pi is 4 pi
  • 00:48:45
    over 3 in 0 to Pi well let's graph it
  • 00:48:48
    really
  • 00:48:51
    quickly now if you recall it's a trick
  • 00:48:53
    we've used plenty of times 4i over 3 can
  • 00:48:56
    be Rewritten it's both a a measurement
  • 00:48:59
    of radians and an arc length on the unit
  • 00:49:01
    circle and for that reason we can
  • 00:49:03
    rewrite it as 4 over3 * pi and it still
  • 00:49:07
    makes sense to us if you look at it this
  • 00:49:09
    way this means that when we go through
  • 00:49:11
    the arc to make the angle 4 pi over 3
  • 00:49:14
    we're going to go through an arc that's
  • 00:49:16
    4/3 the length of Pi or in other words
  • 00:49:18
    slightly bigger than Pi so what we're
  • 00:49:21
    really talking about in fact is an angle
  • 00:49:24
    that falls into quadrant 3
  • 00:49:29
    this right here is 4 pi over
  • 00:49:35
    3 I recommend you double check me when I
  • 00:49:37
    say things like that but you'll quickly
  • 00:49:39
    find that it's the
  • 00:49:41
    truth okay you say this is a problem I
  • 00:49:44
    don't know what the answer is going to
  • 00:49:46
    be then I know that cosine of 4i over 3
  • 00:49:49
    will give us answers down here you can
  • 00:49:51
    actually figure out what this terminal
  • 00:49:52
    point is and I will leave that to you
  • 00:49:54
    guys to think about the hint I would
  • 00:49:57
    give you if you want to use that is to
  • 00:49:59
    look for a reference
  • 00:50:03
    angle and you will notice if you start
  • 00:50:05
    working your calculations here that
  • 00:50:07
    since 4 pi over 3 is bigger than Pi the
  • 00:50:10
    reference angle can be found by taking 4
  • 00:50:12
    pi over 3 and subtracting PI from it
  • 00:50:15
    which will give us pi over
  • 00:50:18
    3 but there's actually an easier way to
  • 00:50:21
    do this than to worry about calculating
  • 00:50:23
    cosine of 4i 3 and then trying to
  • 00:50:26
    calculate cosine inverse of whatever
  • 00:50:28
    number we get see here's something that
  • 00:50:30
    we know and we've talked about many
  • 00:50:32
    times if I have an angle and I reflect
  • 00:50:36
    it across the xaxis like I'm about to do
  • 00:50:39
    right now like
  • 00:50:41
    this even though there are going to be
  • 00:50:44
    some changes we're now no longer below
  • 00:50:46
    the x-axis we're above the x-axis so
  • 00:50:48
    that means that we went from a negative
  • 00:50:50
    yalue for our terminal point to a
  • 00:50:53
    positive y-value for our terminal point
  • 00:50:55
    is there any change to the x coordinate
  • 00:50:58
    no the x coordinate for this blue point
  • 00:51:02
    is exactly the same as the x coordinate
  • 00:51:04
    for this red Point down here in fact if
  • 00:51:08
    I were to start calculating you'd very
  • 00:51:10
    quickly find out that if I take cosine
  • 00:51:14
    of 4 pi over 3 or cosine of whatever
  • 00:51:17
    this blue angle is I will wind up
  • 00:51:20
    getting exactly the same answer so now
  • 00:51:24
    here's the only question
  • 00:51:26
    what is this blue angle well because of
  • 00:51:29
    the way that I've reflected things here
  • 00:51:31
    the blue angle and the red angle should
  • 00:51:33
    have exactly the same reference angle
  • 00:51:35
    which is to say the reference angle here
  • 00:51:38
    should be pi over 3 and I'll bet you
  • 00:51:41
    dollars to Donuts if you realize that
  • 00:51:44
    the angle that I'm interested in is an
  • 00:51:47
    angle in quadrant 2 with a reference
  • 00:51:50
    angle of pi over 3 you will quickly
  • 00:51:52
    figure out the right answer it should be
  • 00:51:55
    that this blue angle is 2 piun
  • 00:52:01
    3 why is that significant because now I
  • 00:52:05
    can rewrite this problem what I'm going
  • 00:52:07
    to do is the following I'm going to
  • 00:52:09
    still write cosine inverse of cosine of
  • 00:52:13
    some value but because I know that
  • 00:52:15
    cosine of 4 piun over 3 and cosine of 2
  • 00:52:19
    piun over 3 give me the same answer they
  • 00:52:22
    correspond to point uh coordinates of
  • 00:52:24
    points with the same x coordinate
  • 00:52:27
    there is no difference between writing
  • 00:52:29
    what I was given and what I've written
  • 00:52:31
    here in
  • 00:52:32
    blue now I'll ask you this is 2 pi over
  • 00:52:36
    3 in the subdomain for cosine is it in
  • 00:52:40
    in other words the the range for R
  • 00:52:43
    cosine you betcha and now the property
  • 00:52:45
    from page to applies these two functions
  • 00:52:48
    undo each other and I'm simply left with
  • 00:52:50
    2 pi over
  • 00:52:55
    3 Let's do at least one more problem
  • 00:52:58
    before we call it quits this video is
  • 00:53:00
    getting plenty long but I want to make
  • 00:53:01
    sure that I have as much information in
  • 00:53:03
    front of you as possible and I want to
  • 00:53:05
    make sure I practice this skill again so
  • 00:53:07
    I have another exercise for us almost
  • 00:53:10
    the same but this time we're calculating
  • 00:53:12
    sign inverse of s of 4 pi over
  • 00:53:15
    3 well I already know that I'm running
  • 00:53:17
    into the same problem here that I did up
  • 00:53:20
    above 4i 3 does not belong to the range
  • 00:53:23
    of Ark sign so I can't simply say the
  • 00:53:26
    answer is 4 pi over 3 it does not work
  • 00:53:28
    that way that means I'm going to have to
  • 00:53:31
    do some rewriting again so here's what
  • 00:53:34
    I'm going to do just like last time I'm
  • 00:53:38
    going to go ahead and draw the angle so
  • 00:53:40
    let's do that really
  • 00:53:43
    quickly okay we have 4 pi over 3 drawn
  • 00:53:47
    again and because we've already talked
  • 00:53:49
    about this I've noted the reference
  • 00:53:51
    angle for 4 pi over 3 is pi over 3 now
  • 00:53:55
    last time what we did is as we said Hey
  • 00:53:57
    listen I'm going to rewrite my equation
  • 00:53:59
    because I happen to know that cosine of
  • 00:54:02
    4i 3 and cosine of 2 pi over 3 are the
  • 00:54:05
    same the reason being they are
  • 00:54:08
    reflections of one another across the x
  • 00:54:10
    axis and an xais reflection does not
  • 00:54:13
    change the x coordinate in other words
  • 00:54:16
    it does not change
  • 00:54:17
    cosine unfortunately we saw in the same
  • 00:54:20
    example that the x-axis Reflections do
  • 00:54:23
    change the value of sign s down here at
  • 00:54:27
    4 piun over 3 is negative sign up here
  • 00:54:29
    at 2 piun over 3 is positive but there
  • 00:54:31
    is another type of reflection we could
  • 00:54:33
    do and some of you may have already
  • 00:54:35
    figured out this is what was going to
  • 00:54:36
    happen if I reflect across the y AIS
  • 00:54:40
    this time notice the x value changes on
  • 00:54:44
    one side it's negative on the other side
  • 00:54:46
    it's positive but does the Y value
  • 00:54:48
    change no it does not symmetry of the
  • 00:54:51
    unit circle tells us this is going to be
  • 00:54:53
    a point that has the same y-coordinate
  • 00:54:55
    as our red point over here we also know
  • 00:54:59
    that when we're talking about sign sign
  • 00:55:02
    is restricted to the right hand side of
  • 00:55:04
    the unit circle so we're in the right
  • 00:55:05
    spot and to get down into Quadrant 4 we
  • 00:55:08
    have to go in the negative
  • 00:55:10
    Direction plus my red angle and my blue
  • 00:55:13
    angle must have exactly the same
  • 00:55:15
    reference angle so doesn't that
  • 00:55:20
    mean that the reference angle here is
  • 00:55:22
    telling us that this angle here is pi
  • 00:55:25
    over 3
  • 00:55:27
    yes it is telling us that so how do we
  • 00:55:30
    rewrite this problem well we write
  • 00:55:33
    exactly the same expression that we were
  • 00:55:35
    given except that we say Hey listen I
  • 00:55:38
    don't need to take s of 4 pi over 3 I
  • 00:55:41
    would get the same answer if I took s of
  • 00:55:44
    pi over
  • 00:55:47
    3 now I've Rewritten the equation it's
  • 00:55:51
    not going to change my answer these two
  • 00:55:52
    things are exactly the same but now that
  • 00:55:55
    I'm looking at at the second equation
  • 00:55:57
    notice that the angle on the inside is
  • 00:56:00
    an angle in the range of Ark sign since
  • 00:56:04
    we are now looking at ARK sign of sign
  • 00:56:07
    of an angle and the angle obeys our
  • 00:56:09
    rules now I can simply let Ark sign and
  • 00:56:12
    sign annihilate each other and so the
  • 00:56:14
    answer becomes piun over
  • 00:56:19
    3 and with that I think that's almost
  • 00:56:22
    all the time we have a couple of quick
  • 00:56:24
    notes for my student students you may be
  • 00:56:27
    wondering uh what other problems there
  • 00:56:29
    might be in this section that you need
  • 00:56:31
    to worry about the only other ones that
  • 00:56:33
    we have not touched are feeding values
  • 00:56:36
    to arc sign Arc cosine Etc on your
  • 00:56:38
    calculator now because of the way this
  • 00:56:41
    video format works out I can't really
  • 00:56:43
    show those to you in person however if
  • 00:56:45
    you have the ti3 x2s first of all notice
  • 00:56:49
    that when you're looking for these
  • 00:56:51
    inverse trig functions you should find
  • 00:56:53
    them by locating first the original trig
  • 00:56:55
    function like s cosine Etc and then
  • 00:56:59
    simply hitting second followed by the
  • 00:57:01
    appropriate trig function so if I press
  • 00:57:03
    the second button on my calculator and
  • 00:57:05
    then sign I should be dealing with arc
  • 00:57:07
    sign now no problems the next thing I
  • 00:57:10
    want to address is that there are more
  • 00:57:12
    identities I mentioned earlier that
  • 00:57:14
    there were equations that I wasn't going
  • 00:57:16
    to be showing because I didn't find them
  • 00:57:18
    to be as useful and I showed you why I
  • 00:57:21
    also have some other ways and some other
  • 00:57:25
    uh relationships we can use you may be
  • 00:57:27
    wondering for example could I do a
  • 00:57:29
    similar trick to what I did in these
  • 00:57:31
    last two examples if I worked with
  • 00:57:33
    Tangent yes there is a trick for working
  • 00:57:36
    with Tangent as well however we've run
  • 00:57:38
    out of time I feel like an hourlong
  • 00:57:40
    video which we pretty much have here is
  • 00:57:42
    plenty long enough instead I will give
  • 00:57:45
    anybody who wants to a chance to ask me
  • 00:57:47
    questions either in the comment section
  • 00:57:48
    of this video or in canvas where USF
  • 00:57:51
    holds its classes for now that's all the
  • 00:57:54
    time I've got so everybody thank you so
  • 00:57:56
    much for your attention thank you for
  • 00:57:58
    your continued hard work if you have
  • 00:58:00
    questions feel free to reach out to me
  • 00:58:02
    and until I see you next time happy math
Etiquetas
  • Trigonometry
  • Inverse functions
  • University of South Florida
  • Mathematics lecture
  • Horizontal line test
  • Domain and range
  • Angles
  • Calculator usage
  • Arc functions
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