Prove that cos10x+cos8x+3cos4x+3cos2x = 8cosxcoscube3x

LHS of the given equation is;
cos10x+cos8x+3cos4x+3cos2x=(cos10x+cos2x)+(cos8x+cos4x)+2(cos4x+cos2x)=2cos6x.cos4x+2cos6x.cos2x+2(cos4x+cos2x) [since cosC+cosD=2cosC+D2.cosC-D2=2cos6x(cos4x+cos2x)+2(cos4x+cos2x)=2(cos4x+cos2x)[cos6x+1]=2.2cos3x.cosx[2cos23x-1+1]    [since cos2θ=2cos2θ-1]=4cos3x.cosx.[2cos23x]=8cosx.cos33x=RHS

hope this helps you

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