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Math Help - More trig questions.

  1. #1
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    More trig questions.

    Once again I am very greatful for you can help.

    1 ) Find the exact value of 17pie / 12

    2) Find the solutions of 2-sin²x = 2cos²(x/2) that are on the interval [0, 2pie)

    For the first, I'm pretty sure I know what to do but I'm simply thinking about it too hard... and the second, im thinking about somehow using the half angle formulas or something like that.
    Last edited by aargh27; December 20th 2006 at 07:15 PM. Reason: (forgot the interval 0,2pie)
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  2. #2
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    Hello, aargh27!

    Is there a typo in #1? .Did you leave out a trig function?
    . . By the way, \pi is spelled "pi".


    1 ) Find the exact value of \frac{17\pi}{12}

    2) Find the solutions of: 2 - \sin^2x \:= \:2\cos^2\left(\frac{x}{2}\right) on the interval [0,\,2\pi)

    Replace \sin^2x with 1 - \cos^2x
    Replace \cos^2\left(\frac{x}{2}\right) with \frac{1 + \cos x}{2}

    Then we have: . 2 - (1 - \cos^2x) \:=\:2\left(\frac{1 + \cos x}{2}\right)

    . . which simplifies to: . \cos^2x - \cos x\:=\:0

    . . which factors: . \cos x(\cos x - 1)\:=\:0


    And we have two equations to solve:

    . . \cos x \:=\:0\quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}

    . . \cos x - 1 \:=\:0\quad\Rightarrow\quad\cos x\:=\:1\quad\Rightarrow\quad\boxed{ x \:=\:0}

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  3. #3
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    Ahh yes.. I did make an error. Sorry about that. Its sin 17
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  4. #4
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    Hello again, aargh27!

    You should know this identity: . \sin^2\theta \:=\:\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad\sin\theta \:=\:\pm\sqrt{\frac{1 - \cos2\theta}{2}}


    1) Find the exact value of \sin\left(\frac{17\pi}{12}\right)

    Note that \frac{17\pi}{12}\:=\:255^o . . . The angle is in Quadrant 3 where \sin is negative.

    . . Using the identity: . \sin\frac{17\pi}{12} \:=\:-\sqrt{\frac{1 - \cos\frac{17\pi}{6}}{2}} [1]

    We have: . \cos\left(\frac{17\pi}{6}\right) \:=\:\cos\left(\frac{5\pi}{6}\right) \:=\:-\frac{\sqrt{3}}{2}


    Substitute into [1]:
    . . \sin\left(\frac{17\pi}{12}\right)\:=\:-\sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{2}\right)}{2}} \:=\:-\sqrt{\frac{2 + \sqrt{3}}{4}} \:=\:-\frac{\sqrt{2 + \sqrt{3}}}{2}

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