3
$\begingroup$

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?

  • 0
    Use the fact $a^3-b^3=(a-b)(a^2+ab+b^2)$.2012-01-25

1 Answers 1

8

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 $1+{1\over2}\sin 2x$.

  • 0
    $f(x)={\cos^3 x-\sin^3 x\over\cos x-\sin x}$ is not defined at $x=n\pi+\pi/4,n\in\mathbb Z$ and thus is not continuous at all $x\in\mathbb R$, while a related function g(x)=\begin{cases}{ \cos^3 x -\sin^3 x\over \cos x-\sin x},&x\neq n\pi+\pi/4,\\\frac{3}{2},&x=n\pi+\pi/4,\end{cases} is continuous at all $x\in\mathbb R$ and equals $1 + \frac{1}{2}\sin 2x$ for all $x\in\mathbb R$.2012-01-27