trig help

**jrose011** · Mar 16th 2009, 07:11 PM

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

how do i solve??

**Soroban** · Mar 16th 2009, 09:42 PM

Hello, jrose011!

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

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

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

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

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

**Grandad** · Mar 17th 2009, 02:21 AM

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:

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

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

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

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

Can you complete it now?

Grandad

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