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OpenStudy (anonymous):
cos^2 x equals 1-sin^2x
OpenStudy (anonymous):
cos^6 x equals (cos^2 x)^3
OpenStudy (anonymous):
So it equals ( 1-sin^2 x)^3
OpenStudy (anonymous):
oh from there what would it be? because when i foil it out it I get stuck somewhere in the middle and my answer is always partly off from what it should be
OpenStudy (anonymous):
What should it be?
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OpenStudy (anonymous):
You're going to have to foil it out
OpenStudy (anonymous):
Srry
OpenStudy (anonymous):
the book doesn't say on this one but for the previous questions like cos^4 x sin ^4 x im supposed to get 1/32(3/4 -cos4x+1/4cos8x) and I don't.. awk
OpenStudy (anonymous):
You should get -sin^6(x)+3 sin^4(x)-3 sin^2(x)+1
OpenStudy (anonymous):
oh in cos still tho
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OpenStudy (anonymous):
We'll I think this is the answer
OpenStudy (anonymous):
Oh ok
OpenStudy (anonymous):
I think this is it then:
1/32 (15 cos(2 x)+6 cos(4 x)+cos(6 x)+10)
OpenStudy (anonymous):
ok ill try foiling again thanks :)
OpenStudy (anonymous):
I just converted everything back to cos using all the trig laws and stuff
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OpenStudy (anonymous):
The answer is 1/32 (15 cos(2 x)+6 cos(4 x)+cos(6 x)+10)