Question

If x = ${30}^{\circ }$, verify that :
$\cos 3x=4{\cos }^{3}x-3\cos x$

Answer 1

Given,

$x=30$

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

$\cos 3x=\cos 3\times 30=\cos 90=0$

$4{\cos }^{3}30-3\cos 30=4\times {\left(\frac{\sqrt{3}}{2}\right)}^3-3\times \frac{\sqrt{3}}{2}=0$

Therefore $\cos 3x=4{\cos }^{3}x-3\cos x$