Writing as a function of only cosine without any powers of cosine

**DarkFlambe** · Dec 9th 2012, 09:42 PM

Hi, can anybody help me on this problem? Thanks in advance ^^

Write Cos^3(2θ) * Sine^2(2θ) as a function of only cosine without any powers of cosine.

**MarkFL** · Dec 9th 2012, 09:55 PM

I would begin with:

$\cos^3(2\theta) \cdot\sin^2(2\theta)=\cos(2\theta) \cdot\left(\sin(2\theta)\cos(2\theta) \right)^2$

Now, apply the double-angle identity for sine to the second factor...what do you get?

**Prove It** · Dec 9th 2012, 09:57 PM

Originally Posted by DarkFlambe

Hi, can anybody help me on this problem? Thanks in advance ^^

Write Cos^3(2θ) * Sine^2(2θ) as a function of only cosine without any powers of cosine.

$\displaystyle \begin{align*} \cos^3{(2\theta)} \sin^2{(2\theta)} &= \cos^3{(2\theta)} \left[ \frac{1}{2} - \frac{1}{2}\cos{(4\theta)} \right] \\ &= \cos{(2\theta)} \cos^2{(2\theta)} \left[ \frac{1}{2} - \frac{1}{2} \cos{(4\theta )} \right] \\ &= \cos{(2\theta)} \left[ \frac{1}{2} + \frac{1}{2} \cos{(4\theta)} \right] \left[ \frac{1}{2} - \frac{1}{2} \cos{(4\theta)} \right] \\ &= \cos{(2\theta)} \left[ \frac{1}{4} - \frac{1}{4}\cos^2{(4\theta)} \right] \\ &= \cos{(2\theta)} \left \{ \frac{1}{4} - \frac{1}{4} \left[ \frac{1}{2} + \frac{1}{2} \cos{(8\theta)} \right] \right \} \\ &= \cos{(2\theta)} \left[ \frac{1}{4} - \frac{1}{8} - \frac{1}{8}\cos{(8\theta)} \right] \\ &= \frac{1}{8}\cos{(2\theta)} - \frac{1}{8}\cos{(2\theta)}\cos{(8\theta)} \end{align*}$

**DarkFlambe** · Dec 12th 2012, 08:39 PM

Thanks! This helped a lot~

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