Question

Write the formula of $\cos 3A$?

Answer 1

Hint : We first find the simplification of the given trigonometry $\cos 3A$ according to the identity $\cos 2A=2{{\cos }^{2}}A-1$ and $\sin \left( 2A \right)=2\sin A\cos A$. We need to simplify the angle as a sum of two angles. We already have the identity of $\cos \left( X+Y \right)=\cos X\cos Y-\sin A\sin B$. We replace the values with $X=2A,Y=A$ to find the final solution.

Complete step-by-step answer :
We need to find the simplified form of $\cos 3A$. This is the multiple angle formula. We know that $\cos 2A=2{{\cos }^{2}}A-1$. We also have $\sin \left( 2A \right)=2\sin A\cos A$.
We need to take the term $\left( 2A+A \right)$ on both sides of the identity $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$.
On the right side we have
$\begin{aligned} & \cos \left( 2A+A \right)=\cos \left( 2A \right)\cos A-\sin A\sin \left( 2A \right) \\\ & \Rightarrow \cos \left( 3A \right)=\left( 2{{\cos }^{2}}A-1 \right)\cos A-\sin A\left( 2\sin A\cos A \right) \\\ \end{aligned}$
We simplify the equation to get
$\begin{aligned} & \cos \left( 3A \right)=\left( 2{{\cos }^{2}}A-1 \right)\cos A-\sin A\left( 2\sin A\cos A \right) \\\ & \Rightarrow \cos \left( 3A \right)=2{{\cos }^{3}}A-\cos A-2{{\sin }^{2}}A\cos A \\\ \end{aligned}$
We now replace the value with ${{\sin }^{2}}A=1-{{\cos }^{2}}A$.
We get
$\begin{aligned} & \cos \left( 3A \right)=2{{\cos }^{3}}A-\cos A-2\left( 1-{{\cos }^{2}}A \right)\cos A \\\ & \Rightarrow \cos \left( 3A \right)=2{{\cos }^{3}}A-\cos A-2\cos A+2{{\cos }^{3}}A \\\ & \Rightarrow \cos \left( 3A \right)=4{{\cos }^{3}}A-3\cos A \\\ \end{aligned}$
Therefore, the formula of $\cos 3A$ is $\cos \left( 3A \right)=4{{\cos }^{3}}A-3\cos A$.

Note : The trigonometric functions of multiple angles are the multiple angle formula. Double and triple angles formulas are there under the multiple angle formulas. Sine, tangent and cosine are the general functions for the multiple angle formula.