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Math Help - Exact values

  1. #1
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    Exact values

    Could someone help me with the following three problems?

    1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).

    2) cos[to the -1 power](cos(4[pi]/3)).

    3) tan(sin[to the -1 power](-4/5)).
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mike1 View Post
    Could someone help me with the following three problems?

    1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
    You are expected to know some special triangles, one of which is the
    30, 60, 90 degree triangle, where if the side opposite the 30 degree angle
    is 1 unit the side opposite the 60 degree angle is sqrt(3), and the other side
    is 2. So using this right triangle we see that sin(30)=sqrt(3)/2, hence if:

    x=asin(-sqrt(3)/2),

    then

    sin(x)=-sqrt(3)/2,

    so:

    sin(-x)=sqrt(3)/2.

    Now there are multiple solutions to this, but we want x to be in the range
    [-90,90], so we want -x=60, or x=-60 degrees, or -pi/3 radian.

    Check: asin(-sqrt(3)/2)~=-1.57, -pi/2~=-1.57

    2) cos[to the -1 power](cos(4[pi]/3)).
    Now 4 pi/3= pi+pi/3, so cos(4 pi/3)=cos(pi)cos(pi/3)-sin(pi)sin(pi/3)=-cos(pi/3)
    Now pi/3 radian is 60 degrees so we are back to our special triangle, and cos(pi/3)=1/2.

    Now acos(-1/2)=pi-pi/3=2pi/3.

    Check acos(cos(4pi/3))~=2.094, 2*pi/3~=2.094

    3) tan(sin[to the -1 power](-4/5)).
    Here we need to think 3-4-5 triangle, first:

    tan(asin(-4/5))=-tan(asin(4/5))

    (draw the triangle identify the angle whose sin is 4/5, then find its
    tangent), so:

    tan(asin(-4/5))=-4/3

    Check: tan(asin(-4/5))~=-1.33.., -4/3~=1.33..

    RonL

    Note: when evaluating the inverse trig functions it is usual to give the answer
    in a fundamental range such that the function has a unique value (the principle value).

    That is evaluating x=asin(y) is not the same as finding x such that sin(x)=y,
    since asin(y) gives a single value, while solving sin(x)=y has multiple solutions.

    For asin the range is usually -pi/2 to pi/2 (or in degrees -90 to 90)
    For acos the range is usually 0 to pi (0 to 180 degrees)
    For atan the range is usually -pi/2 to pi/2 (or sometimes 0 to pi) (-90 to 90 degrees).
    Last edited by CaptainBlack; January 13th 2007 at 03:28 AM. Reason: explaining single valued usage of inverse trig functions
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  3. #3
    MHF Contributor
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    Quote Originally Posted by mike1 View Post
    Could someone help me with the following three problems?

    1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).

    2) cos[to the -1 power](cos(4[pi]/3)).

    3) tan(sin[to the -1 power](-4/5)).
    I assume sin[to the -1 power] is arcsine.

    1) arcsin[-sqrt(3) /2]
    That is an angle whose sine is -sqrt(3) /2. Let's call it theta.

    First, about the negative sign.
    In what quadrants is the sine negative? Since sine is y/r, and r is always positive, then sine is negative wherever y is negative. Below the origin, or below the x-axis. In the 3rd and 4th quadrants.
    So theta is between pi and 2pi radians.

    Then the angle.
    If sin(theta) = sqrt(3) /2, then theta is 60degrees or pi/3 radians. What are the angles in the 3rd and 4th quadrants that are pi/3 from the x-axis?
    Why, (pi +pi/3) in the 3rd quadrant and (2pi -pi/3) in the 4th quadrant.
    And those are 4pi/3 and 5pi/3 radians. -------answer.

    ----------------------
    2) cos[to the -1 power](cos(4[pi]/3)).
    That is "the cosine of an angle whose cosine is 4pi/3".
    Therefore, 4pi/3, of course. ----------answer.

    ======
    Ooppss, my mistake.
    That is actually arccosine of cos(4pi/3).
    An angle whose cosine is cos(4pi/3).

    I do not know how to do that.

    You sure you typed it right?
    -------------------------------------
    3) tan(sin[to the -1 power](-4/5)).
    That is "the tangent of an angle whose sine is -4/5 "

    First, about the negative sine value.
    As in part (i) above, theta is in the 3rd or 4th quadrant, or theta between pi and 2pi radians.

    Then the angle.
    If sin(alpha) = 4/5, then y=4 and r=5 since sine = y/r.
    tan is y/x. So we need to find x.
    By Pythagorean theorem, x = sqrt(r^2 -y^2) = sqrt(5^2 -4^2) = 3.
    Hence tan(alpha) = y/x = 4/3.

    In the 3rd quadrant, x is negative, and y is negative also, hence,
    tan(theta) = y/x = -4/ -3 = 4/3 ------***

    In the 4th quadrant, x is positive, and y is negative, hence,
    tan(theta) = y/x = -4/ 3 = -4/3 -----***

    Therefore, tan[arcsin(-4/5)] = +,-4/3 -----------answer.

    -----------------------
    Note, in the reference triangle,
    opposite side = y
    adjacent side = x
    hypotenuse = r --------always positive.
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  4. #4
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    Hello, Mike!

    ticbol did an excellent job of explaining.
    Here are my versions . . .

    I'll assume the angle are between 0 and 2\pi.


    Find the exact value of:

    (1)\;\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)

    We want the angle whose sine is -\frac{\sqrt{3}}{2}

    You are expected to know that: \sin\left(\frac{\pi}{3}\right) \,=\,\frac{\sqrt{3}}{2}

    Sine is negative in quadrants 3 and 4.
    So we have a reference angle of \frac{\pi}{3} in quadrants 3 and 4.

    Hence, the angles are: . \frac{4\pi}{3},\:\frac{5\pi}{3}



    (2)\;\cos^{-1}\left[\cos\left(\frac{4\pi}{3}\right)\right]

    Inside, we have: . \cos\frac{4\pi}{3} \:=\:\text{-}\frac{1}{2}

    Then we have: . \cos^{-1}\left(\text{-}\frac{1}{2}\right) . . . the angle whose cosine is \text{-}\frac{1}{2}

    We know that: . \cos\left(\frac{\pi}{3}\right) \,=\,\frac{1}{2}
    Cosine is negative in quadrants 2 and 3.

    Therefore the angles are: . \frac{2\pi}{3},\:\frac{4\pi}{3}



    (3)\;\tan\left[\sin^{-1}\left(\text{-}\frac{4}{5}\right)\right]

    Inside, we have: . \sin^{-1}\left(\text{-}\frac{4}{5}\right) . . . the angle whose sine is \text{-}\frac{4}{5}

    Consider: \sin\theta = \frac{4}{5}
    We don't know the exact value of this angle,
    . . but we know it comes from this triangle:
    Code:
                      *
                    * |
                  *   |
              5 *     | 4
              *       |
            * θ       |
          * - - - - - *
                3

    Since sine is negative in quadrants 3 and 4,
    . . the two possible angles look like this:
    Code:
                  |                         |
                  |                         |
             -3   |                         |   3
        - + - - - + - - - - -       - - - - + - - - + -
          :   θ / |                         | \ θ   :
        -4:   /5  |                         |  5\   :-4
          : /     |                         |     \ :
          *       |                         |       *

    Since \tan\theta \:=\:\frac{\text{-}4}{\text{-}3}\text{ or }\frac{\text{-}4}{3}

    . . the answers are: . \pm\frac{4}{3}

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