If sin^4x = A + Bcos2x +Ccos4x, then what is A, B, and C equal to?

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My teacher has yet to discuss any of this, if she even plans on doing so. I'd very much appreciate understanding how to do the problem, so I can do similar ones myself. I don't understand how to handle sin^4x or how to generally isolate the individual variables. Am I just simplifying and reordering to solve for A, B, and C?

Also, how do I handle trigonometric functions with powers above 2? For example, if I know sin^2x=(1-cos2x)/2, what does sin^4x equal? This isn't a problem I need to solve for my work, but I feel like it will help me understand what exactly is going on in these sort of identities.

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Answer 1 · 2017-10-15T06:02:03

# A=3/8, B=-1/2, and, C=1/8.#

We know that, #sin^2x=(1-cos2x)/2, and, cos^2x=(1+cos2x)/2.#

#:. sin^4x=(sin^2x)^2=((1-cos2x)/2)^2,#

#=1/4{1-2cos2x+cos^2 2x},#

#=1/4{1-2cos2x+(cos2x)^2},#

#=1/4{1-2cos2x+(1+cos(2(2x)))/2},#

#=1/8{2-4cos2x+1+cos4x}.#

# rArr sin^4x=3/8-1/2*cos2x+1/8*cos4x.#

Therefore, #sin^4x=A+Bcos2x+Ccos4x,#

# rArr A=3/8, B=-1/2, and, C=1/8.#

Enjoy Maths.!