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1+cos(8x)= A. 4cos(2x) B. 2sin^2(4x) C.4sin(2x) D.2cos^(4x)
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cos 2x=1-2sin^2x
use this
Try using \(\large cos^2x = \frac{1}{2} (1+cos(2x)) \)
but how would I use it?
there is another identity... cos2x=2cos^2x-1
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I'm sorry, I'm really confused
Answer is D, see how....
cos2x=2cos^2x-1 =>1+cos2x=2cos^2x now instead of 2x put 8x(left side) so in the right side its half of 8 that is 4. Got it?
yeah, i think so. thanks!
My pleasure
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