## anonymous 5 years ago use power reduction formula to rewrite the equation in terms of cosine: cos^2x sin^4 x

1. anonymous

Do you have to derive each before you can use them?

2. anonymous

i dont think so

3. anonymous

Okay. Well, $\cos^2 x = \frac{1+\cos 2 x}{2}$and$\sin^4 x = \frac{3-4\cos 2 x +\cos 4 x }{8}$

4. anonymous

You just have to multiply the two.

5. anonymous

All your entries will be in terms of cosine.

6. anonymous

Are you okay to take it from here?

7. anonymous

i get it now thank you very much

8. anonymous

i have trouble multiply these two together, can you help me?

9. anonymous

Yes, just give me a while. I need to eat...starving.

10. anonymous

$\frac{\left(1+\cos 2x \right)}{2}.\frac{\left( 3-4\cos 2x + \cos 4x \right)}{8}$ $=\frac{1}{16}\left(1+\cos 2x \right)\left( 3-4\cos 2x + \cos 4x \right)$$=\frac{1}{16}(3-4\cos 2x + \cos 4x+3\cos 2x -4\cos^2 2x$$+\cos 2x \cos 4x)$$=\frac{1}{16}(3-\cos 2x +\cos 4x (1+ \cos 2x)-4\cos ^2 2x)$

11. anonymous

You can keep going to reduce the square on cos^2(2x) by noting:$\cos^2 \theta -\sin^2 \theta =\cos 2 \theta \rightarrow 2\cos^2 \theta -1= \cos 2 \theta$so$\cos^2 \theta = \frac{\cos 2 \theta +1 }{2}$Use $\theta = 2x$then$\cos^2 \theta = \cos^2 2x = \frac{\cos 4x+1}{2}$which you can substitute back in. Then you have$\frac{1}{16}(3-\cos 2x + \cos 4x (1+\cos 2x)-4.\frac{\cos 4x+1}{2})$$=\frac{1}{16}(1-\cos 2x + \cos 4x (\cos 2x -1))$