If Sin x - cos x = 1/3 , what is sin 3x + cos 3x?

Aug 23, 2016

$- \frac{25}{27}$

Explanation:

$\sin x - \cos x = \frac{1}{3}$

Squaring both sides

$\implies {\left(\sin x - \cos x\right)}^{2} = \frac{1}{9}$

$\implies {\sin}^{2} x + {\cos}^{2} x - 2 \sin x \cos x = \frac{1}{9}$

$\implies 1 - 2 \sin x \cos x = \frac{1}{9}$

$\implies 2 \sin x \cos x = 1 - \frac{1}{9} = \frac{8}{9}$

$\implies \sin x \cos x = \frac{4}{9}$

Now
$\sin 3 x + \cos 3 x$

$= 3 \sin x - 4 {\sin}^{3} x + 4 {\cos}^{3} x - 3 \cos x$

$= 3 \left(\sin x - \cos x\right) - 4 \left({\sin}^{3} x - {\cos}^{3} x\right)$

$= 3 \times \frac{1}{3} - 4 \left(\sin x - \cos x\right) \left({\sin}^{2} x + {\cos}^{2} x + \sin x \cos x\right)$

$= 1 - 4 \times \frac{1}{3} \left(1 + \frac{4}{9}\right)$
$= 1 - \frac{4}{3} \times \frac{13}{9}$
$= 1 - \frac{52}{27} = - \frac{25}{27}$