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
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) $