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Math Help - Trigonometryc identity

  1. #1
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    Trigonometryc identity

    Can you prove that:
    (7+\cos (4x))^2=32[1+\sin^8(x)+\cos^8(x)] ?
    Help needed.
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  2. #2
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    Re: Trigonometryc identity

    Hello, mathmagix!

    \text{Prove that: }\:(7+\cos 4x)^2\:=\:32\left(1+\sin^8x+\cos^8x)
    We will use these identity: . \begin{Bmatrix}\sin^2\theta &=& \dfrac{1-\cos2\theta}{2} \\ \cos^2\!\theta &=& \dfrac{1 + \cos2\theta}{2}\end{Bmatrix}


    The right side is: . 32\left[1 + \sin^8x + \cos^8x\right]

    . . . =\;32\left[1 + (\sin^2x)^4 + (\cos^2x)^4\right] \;=\;32\left[1 + \left(\frac{1-\cos2x}{2}\right)^4 + \left(\frac{1+\cos2x}{2}\right)^4\right]

    . . . =\;32\left[1 + \frac{1 - 4\cos2x + 6\cos^22x - 4\cos^32x + \cos^42x}{16} + \frac{1+4\cos^22x + 6\cos^22x + 4\cos^32x + \cos^42x}{16}\right]

    . . . =\;32\left[1 + \frac{2   + 12\cos^22x + 2\cos^42x}{16}\right] \;=\;32\left[\frac{8 + 1 + 6\cos^22x + \cos^42x}{8}\right]

    . . . =\;4\left[9 + 6\cos^22x + \cos^42x\right] \;=\;4\left[9 + 6\left(\frac{1+\cos4x}{2}\right) + (\cos^22x)^2\right]

    . . . =\;4\left[9 + 3(1 + \cos4x) + \left(\frac{1+\cos4x}{2}\right)^2\right] \;=\;4\left[9 + 3+3\cos4x + \frac{1 + 2\cos4x + \cos^24x}{4}\right]

    . . . =\;4\left[\frac{48 + 12\cos4x + 1 + 2\cos4x + \cos^24x}{4}\right] \;=\;49 + 14\cos4x + \cos^24x

    . . . =\;(7 + \cos4x)^2
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