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Thread: [SOLVED] my last two double angle questions

  1. #1
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    [SOLVED] my last two double angle questions

    I am pretty sure I'm almost done the first one.
    #1
    Use the identity for sin2x and an identity for cos2x to verify that

    but when I use the formula for double angles like where with a
    I think it is

    Thanks for any help you may have.


    And
    #2
    how do you find roots/solve something like
    $\displaystyle 4sinxcosx-1=0$ or $\displaystyle cos^x-sin^2x=1$ I algebraically? I have done questions where you have to do proofs, but I have no examples like this from my teacher, thanks.

    sorry about the multiple questions but the site was down lol.
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  2. #2
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    Quote Originally Posted by brentwoodbc View Post
    I am pretty sure I'm almost done the first one.
    #1
    Use the identity for sin2x and an identity for cos2x to verify that

    but when I use the formula for double angles like where with a
    I think it is

    Thanks for any help you may have.
    $\displaystyle \sin ^2 2x + \cos ^2 2x \hfill \\$

    $\displaystyle = \left( {2\sin x\cos x} \right)^2 + \left( {\cos ^2 x - \sin ^2 x} \right)^2 \hfill \\$

    $\displaystyle = 4\sin ^2 x\cos ^2 x + \cos ^4 x - 2\cos ^2 x\sin ^2 x + \sin ^4 x \hfill \\$

    $\displaystyle = \sin ^4 x + 2\cos ^2 x\sin ^2 x + \cos ^4 x \hfill \\$

    $\displaystyle = \left( {\sin ^2 x + \cos ^2 x} \right)^2 = 1^2 = 1 \hfill \\ $
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  3. #3
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    Quote Originally Posted by brentwoodbc View Post
    #2
    how do you find roots/solve something like
    $\displaystyle 4sinxcosx-1=0$ or $\displaystyle cos^x-sin^2x=1$ I algebraically? I have done questions where you have to do proofs, but I have no examples like this from my teacher, thanks.

    sorry about the multiple questions but the site was down lol.
    $\displaystyle 4 \sin x \cos x - 1 = 0$

    $\displaystyle 2(2\sin x \cos x) - 1 = 0$

    $\displaystyle 2\sin 2x -1=0$

    $\displaystyle \sin 2x =\frac{1}{2}$

    $\displaystyle 2x = \frac{\pi}{6}, \frac{5\pi}{6}$

    $\displaystyle x = \frac{\pi}{12}, \frac{5\pi}{12}$

    Now,

    $\displaystyle \cos x - sin^2 x =1$

    Change $\displaystyle sin^2 x $ into $\displaystyle 1- \cos ^2 x $, then solve the quadratic eqn that is formed.
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  4. #4
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    thanks, very much.
    ya on the first one factoring the
    $\displaystyle = \sin ^4 x + 2\cos ^2 x\sin ^2 x + \cos ^4 x \hfill \\$
    was confusing me because of all the $\displaystyle sin^4$ 's etc.
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