Thread: trig help

express in terms of the first power of cosine: (cos^2(x))(sin^2(x))

how do i solve??

Hello, jrose011!

We're expected to know these Half-angle Identities:

. . $\displaystyle \begin{array}{ccc}\cos^2\!\theta &=& \dfrac{1+\cos2\theta}{2} \\ \\[-4mm] \sin^2\!\theta &=& \dfrac{1-\cos2\theta}{2} \end{array}$

Express in the first power of cosine: .$\displaystyle \cos^2\!x\sin^2\!x$

We have: .$\displaystyle \cos^2\!x\sin^2\!x \;=\; \frac{1+\cos 2x}{2}\cdot\frac{1-\cos2x}{2} \;=\;\frac{1}{4}\left[1 - \cos^2\!2x\right]$

. . . . . . . $\displaystyle = \;\frac{1}{4}\bigg[1 - \frac{1+\cos4x}{2}\bigg] \;=\; \frac{1}{8}\left(1 - \cos 4x\right)$

Hello jrose011

Originally Posted by jrose011

express in terms of the first power of cosine: (cos^2(x))(sin^2(x))

how do i solve??

I assume that this means that you shouldn't have any power higher than 1, so you'll have to use the double-angle identities:

$\displaystyle \cos^2x = \tfrac{1}{2}(1 + \cos 2x)$ and $\displaystyle \sin^2x = \tfrac{1}{2}(1-\cos 2x)$ to get:

$\displaystyle \cos^2x \sin^2x= \tfrac{1}{2}(1 + \cos 2x)\cdot \tfrac{1}{2}(1-\cos 2x)$

$\displaystyle = \tfrac{1}{4}(1-\cos^22x)$

Then do the same again by expressing $\displaystyle \cos^22x$ in terms of $\displaystyle \cos 4x$.

Can you complete it now?

Grandad