Question

If $A=30{}^\circ $ then prove that $3A=4{{\cos }^{3}}A-3\cos A$.

Answer 1

Hint: We will first take the LHS of the given equation and substitute $A=30{}^\circ $ to find its value. Then we will take the right side of the given equation and then substitute $A=30{}^\circ $ to find its value and prove it to be equal to the left side.

Complete step-by-step answer:
Now, we have been given $A=30{}^\circ $.
Now, we have to prove that $3A=4{{\cos }^{3}}A-3\cos A$.
Now, we will take the left side of the equation and substitute the value of $A=30{}^\circ $.
Now, in left side we have,
$\cos 3A$
Now, we will put $A=30{}^\circ $. So, we have,
$\begin{align}
  & \Rightarrow \cos 3\times 30{}^\circ \\
 & \Rightarrow \cos 90{}^\circ \\
\end{align}$
Now, we know that $\cos 90{}^\circ =0$
$\Rightarrow \cos 90{}^\circ =0$
Now, in right hand side we have,
$4{{\cos }^{3}}A-3\cos A$
Now, we will substitute $A=30{}^\circ $
$4{{\cos }^{3}}\left( 30{}^\circ \right)-3\cos \left( 30{}^\circ \right)$
Now, we know that \[\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}\].
\[\begin{align}
  & \Rightarrow 4{{\left( \dfrac{\sqrt{3}}{2} \right)}^{3}}-3\left( \dfrac{\sqrt{3}}{2} \right) \\
 & \Rightarrow 4\times \dfrac{3\sqrt{3}}{8}-\dfrac{3\sqrt{3}}{2} \\
 & \Rightarrow \dfrac{3\sqrt{3}}{2}-\dfrac{3\sqrt{3}}{2} \\
 & \Rightarrow 4{{\cos }^{3}}A-3\cos A=0 \\
\end{align}\]
So, we have LHS = RHS if $A=30{}^\circ $.
Hence, proved that $\cos 3A=4{{\cos }^{3}}A-3\cos A$ if $A=30{}^\circ $.

Note: It is important to note the fact that to solve this question we have used trigonometric identity that,
\[\begin{align}
  & \cos 30{}^\circ =\dfrac{\sqrt{3}}{2} \\
 & \cos 90{}^\circ =0 \\
\end{align}\]
Also, to solve these type of questions it’s important to remember other trigonometric identities like,
\[\begin{align}
  & \sin 30{}^\circ =\cos 60{}^\circ =\dfrac{1}{2} \\
 & \sin 0{}^\circ =\cos 90{}^\circ =0 \\
 & \sin 90{}^\circ =\cos 0{}^\circ =1 \\
\end{align}\]