Question

Rewrite sin^4(x) so that it involves only the first power of cosine.

Answer 1

\[\large\begin{align*}
\sin^4x&=\left(\sin^2x\right)^2\\
&=\left(\frac{1}{2}-\frac{1}{2}\cos2x\right)^2\\
&=\frac{1}{4}-\frac{1}{2}\cos2x+\frac{1}{4}\cos^22x\\
&=\frac{1}{4}-\frac{1}{2}\cos2x+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{2}\cos4x\right)\\
&=\frac{1}{4}-\frac{1}{2}\cos2x+\frac{1}{8}+\frac{1}{8}\cos4x\\
&=\frac{3}{8}-\frac{1}{2}\cos2x+\frac{1}{8}\cos4x\\

\end{align*}\]

Answer 2

Hm, could I have some explanation for the steps you took to get to the solution?

Answer 3

that was clever
in second line
he used the following trig identity (this is how its usually given in the books)

cos 2x = 1 - 2sin^2 x
rearranging we get

sin^2 x = , = 1/2( 1 - cos 2x) or 1/2 - 1/2 cos 2x

Answer 4

then after expanding he still had a term 1/4 cos^2 2x (line 3)

Answer 5

OK?

Answer 6

Yeah, I get it now. :D

Answer 7

then in line 4 he used the identity

cos 2x = 2 cos^2 x - 1
and rearranged to make cos^2 x the subject