Question

Prove the following identity:
 $\cos 3x=4{{\cos }^{3}}x-3\cos x$

Answer 1

Hint: Split 3x to 2x and x and apply the identity of $\cos \left( A+B \right)$ to simplify it, which is given as $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B.$ And use the identities of $\sin 2x,\cos 2x$ to solve it further which are given as $\sin 2x=2\sin x\cos x,\cos 2x=2{{\cos }^{2}}x-1$ use the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1,{{\sin }^{2}}x=1-{{\cos }^{2}}x$ to convert any sin function to cos wherever required.

Complete step-by-step answer:
We have to prove
$\cos 3x=4{{\cos }^{3}}x-3\cos x............\left( i \right)$
So, we can simplify the left hand side of the above expression to get the right hand side. So, let
$LHS=\cos 3x..............\left( ii \right)$
Now, we can split the $\angle 3x$ in sum of $\angle 2x,x$ and hence, we get equation (ii) as
$LHS=\cos \left( 2x+x \right)$
Here, we need to apply the trigonometric identity of $\cos \left( A+B \right)$ by taking A = 2x and B = x. So, identity is given as
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B...............\left( iii \right)$
So, $\cos \left( 2x+x \right)$ can be simplified by equation (iii) as
$\begin{align}
  & LHS=\cos \left( 2x+x \right)=\cos 2x\cos x-\sin 2x\sin x \\
 & LHS=\cos 2x\cos x-\sin 2x\sin x................\left( iv \right) \\
\end{align}$
Now, we can use trigonometric identities of $\cos 2x,\sin 2x$ to simplify the equation (iv) further. So, $\cos 2x,\sin 2x$ are given as
$\begin{align}
  & \cos 2x=2{{\cos }^{2}}x-1 \\
 & \sin 2x=2\sin x\cos x \\
\end{align}$
Hence, putting the values of $\cos 2x,\sin 2x$ to the expression (iv). So, we get
$\begin{align}
  & LHS=\left( 2{{\cos }^{2}}x-1 \right)\cos x-\left( 2\sin x\cos x \right)\sin x \\
 & LHS=2{{\cos }^{3}}x-\cos x-2{{\sin }^{2}}x\cos x............\left( v \right) \\
\end{align}$
Now, as the RHS of equation (i) is given only in cos functions. So, we need to replace ${{\sin }^{2}}x$ from equation (v) with the help of the trigonometric identity given as
$\begin{align}
  & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
 & {{\sin }^{2}}x=\left( 1-{{\cos }^{2}}x \right)................(vi) \\
\end{align}$
Hence, replacing ${{\sin }^{2}}x$ by $1-{{\cos }^{2}}x$ through the identity of equation (vi) of the expression given in equation (v). So, we get
$\begin{align}
  & LHS=2{{\cos }^{3}}x-\cos x-2\left( 1-{{\cos }^{2}}x \right)\cos x \\
 & LHS=2{{\cos }^{3}}x-\cos x-2\cos x+2{{\cos }^{3}}x \\
 & LHS=4{{\cos }^{3}}x-3\cos x \\
\end{align}$
Hence, we get $LHS=RHS=4{{\cos }^{3}}x-3\cos x$
So, the equation given in the problem is proved.

Note: One may split 3x as $\dfrac{3x}{2},\dfrac{3x}{2}$ and apply the formula of $\cos \left( A+B \right)$, which is the wrong approach or may be complex for further calculations. So, don’t split 3x in the form of fractions.
One may split 3x as (4x – x) as well and can solve it with the help of the trigonometric identities of
$\begin{align}
  & \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\
 & \cos 2x=2{{\cos }^{2}}x-1 \\
 & \sin 2x=2\sin x\cos x \\
 & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
\end{align}$
So, it can be another approach for the question.