Prove the identity, `cos((3pi)/2+4alpha)+sin(3pi-8alpha)-sin(4pi-12alpha)=4cos2alpha cos 4alpha sin 6 alpha.`

Prove that sin^2alpha-sin^4alpha= cos^2alpha-cos^4alpha

(sin 5 alpha-sin3alpha)/(cos 5 alpha+2 cos 4 alpha+cos 3alpha)=

sin alpha-sin2 alpha+sin3 alpha=4cos(3 alpha)/(2)cos alpha sin(alpha)/(2)

cos^(4)alpha+sin^(4)alpha-6sin^(2)alpha cos^(2)alpha=

sin alpha+sin2 alpha+sin4 alpha+sin5 alpha=4cos((alpha)/(2))cos(3(alpha)/(2))sin3 alpha

sin alpha+sin2 alpha+sin4 alpha+sin5 alpha=4cos((alpha)/(2))cos(3(alpha)/(2))sin3 alpha

3[sin^4((3pi)/2-alpha)+sin^4(3pi+alpha)]-2[sin^6(pi/2+alpha)+sin^6(5pi-alpha)]=

(cos2 alpha)/(cos^(4)alpha-sin^(4)alpha)-(cos^(4)alpha+sin^(4)alpha)/(2-sin^(2)2 alpha)=

Prove that, 1/3 ( cos^(3) alpha sin 3 alpha + sin^(3) alpha cos 3 alpha) = 1/4 sin 4 alpha

Prove that (sin alpha+sin2 alpha+sin3alpha+sin4alpha)/(cos alpha+cos 2 alpha+cos 3 alpha+cos4 alpha)=tan((5 alpha)/2)