What is #4cos^5thetasin^5theta# in terms of non-exponential trigonometric functions?

1 Answer
Topscooter
Dec 28, 2015

#1/8sin(2theta)(3-4cos(4theta)+cos(8theta))#

Explanation:

We know that #sin(2x) = 2sin(x)cos(x)#. We apply this formula here!

#4cos^5(theta)sin^5(theta) = 4(sin(theta)cos(theta))^5 = 4(sin(2theta)/2)^5 = sin^5(2theta)/8#.

We also know that #sin^2(theta) = (1-cos(2theta))/2# and #cos^2(theta) = (1+cos(2theta))/2#.

So #sin^5(2theta)/8 = sin(2theta)/8*((1-cos(4theta))/2)^2 = sin(2theta)/8*(1 - 2cos(4theta)+cos^2(4theta))/4 = sin(2theta)/8*((1-2cos(4theta))/4 + (1+cos(8theta))/8) = 1/8sin(2theta)(3-4cos(4theta)+cos(8theta))#

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