Question #6d8e6

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Answer 1 · 2016-01-14T08:39:34

We will make repeated use of the power reduction formulas:

Proceeding,

#sin^2thetacos^2theta = (1 - cos2theta)/2*(1+cos2theta)/2#

#= 1/4(1-cos^2 2theta)#

#= 1/4(1 -(1+cos(2(2theta))/2))#

#= 1/4(1-(1+cos4theta)/2)#

#= 1/4((1-cos4theta)/2)#

#= 1/8(1-cos4theta)#