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Thread: Trig iden: Quadruple angle formulas

  1. #1
    Junior Member
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    Trig iden: Quadruple angle formulas

    How do you tackle these?
    1. 2cos^2 4x-1
    sin4xcos4x





    2. 1-4sin^2x+4sin^4x

    The answers to the first is 2cot8x and the second is cos^22x. I would like to see the step by step process.
    Help would be appreciated thanks!
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Is it:

    1. $\displaystyle \frac{2 \cos^{2} 4x-1}{\sin 4x \cos 4x} $ and

    2. $\displaystyle 1 - 4 \sin^{2}x + 4 \sin^{4} x $?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    We use the addition formulas for sine and cosine here.

    Recall:

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

    $\displaystyle \Rightarrow \cos 8x = 2 \cos^2 4x - 1$

    $\displaystyle \sin 2x = 2 \sin x \cos x$

    $\displaystyle \Rightarrow \sin 8x = 2 \sin 4x \cos 4x$

    Quote Originally Posted by andrewsx View Post
    How do you tackle these?
    1. 2cos^2 4x-1
    sin4xcos4x
    $\displaystyle \frac {2 \cos^2 4x - 1}{\sin 4x \cos 4x} $

    $\displaystyle = \frac {\cos 8x}{ \frac {1}{2} \sin 8x}$

    $\displaystyle = 2 \frac {\cos 8x }{\sin 8x}$

    $\displaystyle = 2 \cot 8x$


    2. 1-4sin^2x+4sin^4x
    $\displaystyle 1 - 4 \sin^2 x + 4 \sin^4 x$ ............Note that this is quadratic in $\displaystyle \sin^2 x$, so let's foil

    $\displaystyle = \left( 1 - 2 \sin^2 x \right) \left(1 - 2 \sin^2 x \right)$

    $\displaystyle = \left( 1 - 2 \sin^2 x \right)^2$

    $\displaystyle = \cos^2 2x$
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