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

Given:

\( \theta=30^{\circ} \)

To do:

We have to verify that \( \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \).

Solution:  

\( \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \)

This implies,

\( \cos 3(30^{\circ})=4 \cos ^{3} 30^{\circ}-3 \cos 30^{\circ} \)

\( \cos 90^{\circ}=4 \cos ^{3} 30^{\circ}-3 \cos 30^{\circ} \)

We know that,

$\cos 90^{\circ}=0$

$\cos 30^{\circ}=\frac{\sqrt3}{2}$

Let us consider LHS,

$\cos 3 \theta=\cos 90^{\circ}$

$=0$

Let us consider RHS,

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

$=4\left(\frac{3\sqrt{3}}{8}\right) -\frac{3\sqrt{3}}{2}$

$=\frac{3\sqrt{3}}{2} -\frac{3\sqrt{3}}{2}$

$=0$

LHS $=$ RHS

Hence proved.

Tutorialspoint

Updated on: 10-Oct-2022

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