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

Question

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.

9230 views around the world

You can reuse this answer
Creative Commons License

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.!