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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you have to derive each before you can use them?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. Well, \[\cos^2 x = \frac{1+\cos 2 x}{2}\]and\[\sin^4 x = \frac{34\cos 2 x +\cos 4 x }{8}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just have to multiply the two.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0All your entries will be in terms of cosine.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you okay to take it from here?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get it now thank you very much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have trouble multiply these two together, can you help me?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, just give me a while. I need to eat...starving.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\left(1+\cos 2x \right)}{2}.\frac{\left( 34\cos 2x + \cos 4x \right)}{8}\] \[=\frac{1}{16}\left(1+\cos 2x \right)\left( 34\cos 2x + \cos 4x \right)\]\[=\frac{1}{16}(34\cos 2x + \cos 4x+3\cos 2x 4\cos^2 2x\]\[+\cos 2x \cos 4x)\]\[=\frac{1}{16}(3\cos 2x +\cos 4x (1+ \cos 2x)4\cos ^2 2x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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\[\frac{1}{16}(3\cos 2x + \cos 4x (1+\cos 2x)4.\frac{\cos 4x+1}{2})\]\[=\frac{1}{16}(1\cos 2x + \cos 4x (\cos 2x 1))\]
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