Question

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

Answer 1

Hint: From the question given we have to prove that $\cos 3x=4{{\cos }^{3}}x-3\cos x$. To solve this question, we have to use the simple trigonometric formulas. First, we have to take $\cos 3x$ . We have to split the $3x$ as $2x+x$ and then we have to apply the $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$. After that we have to apply the formulas like $\cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$. By further simplifying we will get the right-hand side part. Hence the equation will be proved.

Complete step by step answer:
From the question given we have to prove that
$\Rightarrow \cos 3x=4{{\cos }^{3}}x-3\cos x$
First, we have to take the left-hand side part
The left-hand side part is
$\Rightarrow \cos 3x$
Now we have to split the $3x$ as $2x+x$.
By splitting we will get,
$\Rightarrow \cos \left( 2x+x \right)$
As we know that from basic trigonometric identities
$\Rightarrow \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
By this formula we can expand the above part
$\Rightarrow \cos \left( 2x+x \right)=\cos 2x\cos x-\sin 2x\sin x$
Again, from the basic formulas of trigonometric identities the above equation can be expanded as
$\Rightarrow \cos 2x=2{{\cos }^{2}}x-1=1-2{{\sin }^{2}}x$
$\Rightarrow \sin 2x=2\sin x\cos x$
By these formulae we will expand the above part
$\Rightarrow \cos \left( 2x+x \right)=\cos 2x\cos x-\sin 2x\sin x$
$\Rightarrow \cos \left( 2x+x \right)=\left( 2{{\cos }^{2}}x-1 \right)\cos x-\left( 2\sin x\cos x \right)\sin x$
By further simplifying we will get,
$\Rightarrow \cos \left( 2x+x \right)=2{{\cos }^{3}}x-\cos x-2{{\sin }^{2}}x\cos x$
 Again, from the basic formulas of trigonometric identities we know that
$\Rightarrow {{\sin }^{2}}x=1-{{\cos }^{2}}x$
By these formulae we will expand the above part
$\Rightarrow \cos 3x=2{{\cos }^{3}}x-\cos x-2\cos x\left( 1-{{\cos }^{2}}x \right)$
 By further simplifying we will get,
$\Rightarrow \cos 3x=2{{\cos }^{3}}x-\cos x-2\cos x+2{{\cos }^{3}}x$
 By further simplifying we will get,
$\Rightarrow \cos 3x=4{{\cos }^{3}}x-3\cos x$
Therefore, the right hand side part is equal to the left-hand side.

Note: Students should recall the basic formulas of trigonometry and its identities while doing the above problem. Students should know the formulas like,
$\begin{align}
  & \Rightarrow \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\
 & \Rightarrow \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\
 & \Rightarrow \sin \left( A-B \right)=\cos A\sin B-\sin A\cos B \\
 & \Rightarrow \sin \left( A+B \right)=\cos A\sin B+\sin A\cos B \\
\end{align}$