Question

use power reduction formula to rewrite the equation in terms of cosine: cos^2x sin^4 x

Answer 1

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

Answer 2

i dont think so

Answer 3

Okay.

Well, \[\cos^2 x = \frac{1+\cos 2 x}{2}\]and\[\sin^4 x = \frac{3-4\cos 2 x +\cos 4 x }{8}\]

Answer 4

You just have to multiply the two.

Answer 5

All your entries will be in terms of cosine.

Answer 6

Are you okay to take it from here?

Answer 7

i get it now thank you very much

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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\[\frac{1}{16}(3-\cos 2x + \cos 4x (1+\cos 2x)-4.\frac{\cos 4x+1}{2})\]\[=\frac{1}{16}(1-\cos 2x + \cos 4x (\cos 2x -1))\]