# lowering powers

#### reino17

rewrite the expression in terms of the first power of cosine

sin^4x

so far i have
(sin^2x)^2

= (1-cos2x/2)^2

= (1-cos2x/2)(1-cos2x/2)

=1-cos^2.2x/4

so my question is, is it right? or is there more

#### Soroban

MHF Hall of Honor
Hello, reino17!

Your algebra is off . . .

Rewrite the expression in terms of the first power of cosine: .$$\displaystyle \sin^4x$$

$$\displaystyle \sin^4\!x \;\;=\;\;\left(\sin^2\!x\right)^2 \;\;=\;\;\left(\frac{1-\cos2x}{2}\right)^2 \;\;=$$ . $$\displaystyle \frac{1-2\cos2x + \cos^2\!2x}{4}\;\;=\;\;\frac{1 - 2\cos2x + \frac{1 + \cos4x}{2}}{4}$$

Multiply by $$\displaystyle \tfrac{2}{2}:\quad \frac{2-4\cos2x + 1 + \cos4x}{8} \;\;=\;\;\tfrac{1}{8}(3 - 4\cos 2x + \cos4x)$$

reino17

#### reino17

ahh ok i see where i screwed up. thanks dude