Trigonometryc identity

**mathmagix** · August 30th 2012, 02:03 AM

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

**Soroban** · August 30th 2012, 03:51 PM

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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