Prove $\frac{\cos^3{x}-\sin^3{x}}{\cos{x}-\sin{x}} =1+\frac{1}{2} \sin{2x}$ Prove 
$$\frac{\cos^3{x}-\sin^3{x}}{\cos{x}-\sin{x}} =1+\frac{1}{2} \sin{2x}$$
How do I start :( which identity do I use?

Hint: use the identity:
$(a^3-b^3)=(a-b)(a^2+ab+b^2)$.



Solution follows:


We have 
$\cos^3 x -\sin^3 x =(\cos x-\sin x)(\cos^2 x +\cos x\sin x +\sin^2 x )$.
So:
$$\eqalign{{
\cos^3 x -\sin^3 x\over \cos x-\sin x}&= 
{(\cos x-\sin x)(\cos^2 x +\cos x\sin x +\sin^2 x )\over (\cos x-\sin x)}\cr
&=1+\cos x\sin x\cr
&=1+\textstyle{1\over2}\sin 2x.}
$$
(The last equality used the trigonometric identity $\sin(2x)=2\sin x\cos x$.)
 
As @Dilip  Sarwate points out in the comment below, the above does not hold when $\cos x=\sin x$. In this case, ${
\cos^3 x -\sin^3 x\over \cos x-\sin x}$ is not defined.  Of course, whenever  ${
\cos^3 x -\sin^3 x\over \cos x-\sin x}$ is defined, it is equal to $1+{1\over2}\sin 2x$.
