Question

Prove that the following equation is an identity:

(six^3y-cos^3y)/siny-cosy = (2+sin2y)/2

Answer 1

\[\frac{\sin ^3y-\cos ^3y}{\sin y-\cos y}=\frac{(\sin y-\cos y)(\sin ^2y+\sin y \cos y+\cos ^2y)}{\sin y-\cos y}\]=
\[\frac{(\sin y-\cos y)(1+\sin y \cos y)}{\sin y-\cos y}=(1+\sin y \cos y)\times\frac{2}{2}=\]
\[\frac{2+2\sin y \cos y}{2}=\frac{2+\sin 2y}{2}\]

Answer 2

Thanks.

Answer 3

yw