Sin^4x=1/8(3-4cos2x+cos4x) How to prove?

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Answer 1 · 2018-03-09T16:43:34

#LHS=sin^4x#

#=[(2sin^2x)/2]^2#

#=1/4[(1-cos2x)^2]#

#=1/4[1-2cos2x+cos^2(2x)]#

#=2/(4*2)[1-2cos2x+cos^2(2x)]#

#=1/8[2-4cos2x+2cos^2(2x)]#

#=1/8[2-4cos2x+1+cos4x]#

#=1/8[3-4cos2x+cos4x]=RHS#