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Math Help - using the half angle formula correctly

  1. #1
    Junior Member
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    using the half angle formula correctly

    Here is the problem- If cos(t) = -9/11 where pi<t<3pi/2, find the values of the following trigonometric functions. the answer has to be a fraction or an arithmetic expression, exact answers.
    for:
    cos(2t)
    sin(2t)
    cos(t/2)
    sin(t/2)
    Here is what I did. Given that cos(t)= -9/11
    (opp)^2+(adj)^2=(hyp)^2
    (opp)^2 + (-9)^2 + (11)^2
    (opp)^2 =202
    opp = -sqrt(202)

    I also thought since cos(2t) = 2cos^2(x)-1 I could do this
    -9/11 = 2cos^2(x)-1
    2cos^2(x)=2/11
    cos^2(x)=1/11
    cos(x)=squrt(1/11) and the answer in the back of the book tells me I am wrong. If you could get me on track I would really appreciate it. I will reread the chapter and the section on this and maybe it will come to me.
    Thank you for your time!!
    Keith Stevens
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Keith!

    If \cos\theta  = -\frac{9}{11}, where \pi < \theta < \frac{3\pi}{2},
    find the values of the following trigonometric functions.

    (1)\;\cos2\theta\quad(2)\;\sin2\theta\quad(3)\;\co  s\left(\frac{\theta}{2}\right)\quad (4)\;\sin\left(\frac{\theta}{2}\right)

    Here is what I did.
    Given that: \cos\theta = -\frac{9}{11}
    . . (opp)^2 + (adj)^2\:=\:(hyp)^2
    . . (opp)^2 + (-9)^2 \:=\:(11)^2
    . . (opp)^2 \:=\:202 . . . no

    You had: . (opp)^2 + 81 \;=\;121\quad\Rightarrow\quad (opp)^2\,=\,40\quad\Rightarrow\quad opp \,= \,\pm2\sqrt{10}

    Since \theta is in Quadrant 3: opp = -2\sqrt{20}

    So we have: . \begin{array}{ccc}\sin\theta \:= \\ \\ \cos\theta \:=\end{array}\!\!<br />
\begin{array}{ccc}-\frac{2\sqrt{10}}{11}\\ \\ -\frac{9}{11}\end{array}


    (1)\;\cos2\theta\;=\;2\cos^2\theta-1 \;=\;2\left(-\frac{9}{11}\right)^2 - 1\;=\;\frac{162}{121} - 1\;=\;\frac{41}{121}

    (2)\;\sin2\theta\;=\;2\sin\theta\cos\theta \;=\;2\left(-\frac{2\sqrt{10}}{11}\right)\left(-\frac{9}{11}\right)\;=\;\frac{36\sqrt{10}}{121}

    (3)\;\cos\left(\frac{\theta}{2}\right)\;=\;\pm\sqr  t{\frac{1 + \cos\theta}{2}} \;=\;\pm\sqrt{\frac{1 + \left(-\frac{9}{11}\right)}{2}} \;=\;\boxed{-\sqrt{\frac{1}{11}}} *

    (4)\;\sin\left(\frac{\theta}{2}\right)\;=\;\pm\sqr  t{\frac{1 - \cos\theta}{2}} \;=\;\pm\sqrt{\frac{1 - \left(-\frac{9}{11}\right)}{2}}\;=\;\boxed{\sqrt{\frac{10  }{11}}} *

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    *
    Since \cos\theta = -\frac{9}{11}, then \theta \,\approx\,215^o

    Hence: \frac{\theta}{2} \,\approx\,107.5^o is in Quadrant 2.

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