Question

If $\theta ={30}^{\circ },$ verify the following:
$i)\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

$ii)\sin 3\theta =3\sin \theta -4{\sin }^3\theta$

Answer 1

$i)\cos 3\theta =4{\cos }^3\theta -3cos\theta$

$\begin{array}{cc}LHS& =\cos 3\theta \\ & =\cos \left(3\times {30}^{\circ }\right)\\ & =\cos \left({90}^{\circ }\right)\\ & =0\end{array}$

$\begin{array}{cc}RHS& =4{\cos }^3\theta -3\cos \theta \\ & =4{\cos }^3{30}^{\circ }-3\cos {30}^{\circ }\\ & =4{\left(\frac{\sqrt{3}}{2}\right)}^3-3\times \frac{\sqrt{3}}{2}\\ & =\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}\\ & =0\end{array}$

Hence proved, $LHS=RHS$.

$ii)\sin 3\theta =3\sin \theta -4{\sin }^3\theta$

$\begin{array}{cc}LHS& =\sin \left(3\times {30}^{\circ }\right)\\ & =\sin \left({90}^{\circ }\right)\\ & =1\end{array}$

$\begin{array}{cc}RHS& =3\sin \left({30}^{\circ }\right)-4{\sin }^3\left({30}^{\circ }\right)\\ & =3\times \frac{1}{2}-4{\left(\frac{1}{2}\right)}^3\\ & =\frac{3}{2}-\frac{1}{2}\\ & =1\end{array}$

Hnece proved, $LHS=RHS$.