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Can someone please show me how to do this problem?

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Use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine

Sin^4x

ANSWER: 1/2(3/4 - cos 3x + 1/4 cos 4x)
asked Mar 31, 2013 in TRIGONOMETRY by andrew Scholar

1 Answer

+1 vote
sin^4 x = (sin^2 x)^2

Now, sin^2x + cos^2 x=1

so,

sin^4 x = (sin^2 x)^2= (1-cos^2 x)^2

cos 2x = 2cos^2 x-1, so cos^2 x = (1+cos 2x)/2

sin^4 x = (sin^2 x)^2= (1-cos^2 x)^2 = (1- (1+cos 2x)/2 )^2 = (1/2- cos 2x/2)^2 = 1/4 (1-cos 2x)^2=1/4 (1-2cos 2x+cos^22x)

again using, cos 2x = 2cos^2 x-1, so cos^2 2x = (1+cos 4x)/2

sin^4x = 1/4(1-2cos2x + (1+cos4x)/2) = 1/4(3/2 -2 cos 2x +cos 4x/2) = =(1/2)(3/4-cos 2x +1/4 cos 4x)

The answer you mentioned must have cos 2x instead of cos 3x
answered Mar 31, 2013 by hartnn Rookie
Thanks!!!!!!!!!!!!!!

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