precalculus
posted by lina .
Rewrite the expression so that it involves the sum or difference of only constants and sines and cosine to the 1st power
(sinx)^4

sin^4
(1cos^2)^2
(1cos)^2(1+cos)^2
(1cos)(1cos)(1+cos)(1+cos)
sin^2(x) = (1cos(2x))/2
so,
sin^4(x) = (1  cos(2x))^2/4
= (1  2cos(2x) + cos^2(2x))/4
= (1  2cos(2x) + (1  cos(4x))/2)/4
= (1  2cos(2x) + 1/2  cos(4x)/2)/4
= (2  4cos(2x) + 1  cos(4x))/8
= (3  4cos(2x)  cos(4x))/8
don't know which way you wanted to go there.
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