# Math Help - lowering powers

1. ## lowering powers

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

2. Hello, reino17!

Your algebra is off . . .

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

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

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

3. ahh ok i see where i screwed up. thanks dude