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Math Help - [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
    4sinxcosx-1=0 or 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.
     \sin ^2 2x + \cos ^2 2x \hfill \\

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

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

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

    = \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
    4sinxcosx-1=0 or 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.
    4 \sin x \cos x - 1 = 0

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

    2\sin 2x -1=0

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

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

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

    Now,

    \cos x - sin^2 x =1

    Change sin^2 x into 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
    = \sin ^4 x + 2\cos ^2 x\sin ^2 x + \cos ^4 x \hfill \\
    was confusing me because of all the sin^4 's etc.
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