Question

Prove-: sin5A = 5cos^4 A sinA - 10cos^2 A sin^3 A + sin^5 A

Answer 1

$A=x\\\\sin5x=5cos^4xsinx-10cos^2xsin^3x+sin^5x\\\\L=sin(3x+2x)=sin3xcos2x+sin2xcos3x=(*)\\-------------------------------\\sin3x=sin(2x+x)=sin2xcosx+sinxcos2x\\=2sinxcosxcosx+sinx(cos^2x-sin^2x)\\=2sinxcos^2x+sinxcos^2x-sin^3x=3sinxcos^2x-sin^3x\\-------------------------------$

$sin3xcos2x=(3sinxcos^2x-sin^3x)(cos^2x-sin^2x)\\=3sinxcos^4x-3sin^3xcos^2x-sin^3xcos^2x+sin^5x\\=3sinxcos^4x-4sin^3xcos^2x+sin^5x\\-------------------------------$

$cos3x=cos(2x+x)=cos2xcosx-sin2xsinx\\=cosx(cos^2x-sin^2x)-2sinxcosxsinx\\=cos^3x-sin^2xcosx-2sin^2xcosx=cos^3x-3sin^2xcosx\\-------------------------------$

$sin2xcos3x=2sinxcosx(cos^3x-3sin^2xcosx)\\=2sinxcos^4x-6sin^3xcos^2x\\-------------------------------$

$(*)=3sinxcos^4x-4sin^3xcos^2x+sin^5x+2sinxcos^4x-6sin^3xcos^2x\\\\=5sinxcos^4x-10sin^3xcos^2x+sin^5x=R$

$=======================================\\\\sin(\alpha+\beta)=sin\alpha cos\beta+sin\beta cos\alpha\\\\cos(\alpha+\beta)=cos\alpha cos\beta-sin\alpha sin\beta\\\\sin2\alpha=2sin\alpha cos\alpha\\\\cos\alpha=cos^2\alpha-sin^2\alpha$