• Apr 17th 2013, 04:10 AM
aser03
Use the formulas for lowering powers to rewrite: cos^4x sin^4x in terms of the first power of cosine?
help me please I dont know how to do it.
• Apr 17th 2013, 05:51 AM
Soroban
Hello, aser03!

We are expected to know these identities:

. . $\sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin^2\!\theta \:=\:1-\cos^2\!\theta$

. . $\sin2\theta \,=\,2\sin\theta\cos\theta$

. . $\cos^2\!\theta \:=\:\frac{1+\cos2\theta}{2}$

Quote:

$\text{Use the formulas for lowering powers to rewrite }\cos^4\!x\sin^4\!x$
. . $\text{ in terms of the first power of cosine.}$

$\cos^4\!x\sin^4\!x \;=\;(\sin x\cos x)^4$

. . . . . . . . $=\;\tfrac{1}{16}(2\sin x\cos x)^4$

. . . . . . . . $=\;\tfrac{1}{16}(\sin2x)^4$

. . . . . . . . $=\;\tfrac{1}{16}(\sin^2\!2x)^2$

. . . . . . . . $=\;\tfrac{1}{16}(1-\cos^2\!2x)^2$

. . . . . . . . $=\;\tfrac{1}{16}\left(1 - \tfrac{1\:+\:\cos4x}{2}\right)^2$

. . . . . . . . $=\;\tfrac{1}{16}\left(\tfrac{2\,-\,(1\,+\,\cos4x)}{2}\right)^2$

. . . . . . . . $=\;\tfrac{1}{16}\left(\tfrac{1\,-\,\cos4x}{2}\right)^2$

. . . . . . . . $=\;\tfrac{1}{16}\cdot\tfrac{1}{4}(1-\cos4x)^2$

. . . . . . . . $=\;\tfrac{1}{64}(1 - 2\cos4x + \cos^2\!4x)$

. . . . . . . . $=\;\tfrac{1}{64}\left(1 - 2\cos4x + \tfrac{1 + \cos8x}{2}\right)$

. . . . . . . . $=\;\tfrac{1}{64}\left(1 - 2\cos4x + \tfrac{1}{2} + \tfrac{1}{2}\cos8x\right)$

. . . . . . . . $=\;\tfrac{1}{64}\left(\tfrac{3}{2} - 2\cos4x + \tfrac{1}{2}\cos8x\right)$

. . . . . . . . $=\;\tfrac{1}{128}(3 - 4\cos4x + \cos8x)$