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?
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:
$(a^3-b^3)=(a-b)(a^2+ab+b^2)$.
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 $\sin(2x)=2\sin x\cos x$.)