Question

Prove that:-
\[2\cos x-\cos 3x-\cos 5x=16{{\cos }^{3}}x{{\sin }^{2}}x\]

Answer 1

Hint: In such questions, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.
The below mentioned formulae may be used before solving, in the solution which is as follows
\[\begin{align}
  & \Rightarrow \cos (2\theta )=1-2{{\sin }^{2}}\theta \\
 & \Rightarrow \cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) \\
 & \Rightarrow \sin 2x=2\sin x\cdot \cos x \\
\end{align}\]

Complete step-by-step answer:

As mentioned in the question, we have to prove the given expression.
Now, we will start with the left hand side that is L.H.S. and try to make the necessary changes that are given in the hint, first, as follows
\[\begin{align}
  & \Rightarrow 2\cos x-\cos 3x-\cos 5x \\
 & \Rightarrow 2\cos x-\left( \cos 3x+\cos 5x \right) \\
 & \left[ \cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right) \right] \\
 & \Rightarrow 2\cos x-\left( 2\cos \left( \dfrac{3x+5x}{2} \right)\cos \left( \dfrac{3x-5x}{2} \right) \right) \\
 & \Rightarrow 2\cos x-\left( 2\cos \left( 4x \right)\cos \left( -x \right) \right) \\
 & \left[ \cos (-x)=\cos x \right] \\
 & \Rightarrow 2\cos x-\left( 2\cos \left( 4x \right)\cos \left( x \right) \right) \\
 & \Rightarrow 2\cos x\left( 1-\left( \cos \left( 4x \right) \right) \right) \\
 & \left[ \cos 2x=1-2{{\sin }^{2}}x \right] \\
 & \Rightarrow 2\cos x\left( 2{{\sin }^{2}}2x \right) \\
 & \Rightarrow 4\cos x\left( {{\sin }^{2}}2x \right) \\
\end{align}\]
Now, using the relation given in the hint, we can write as follows
\[\begin{align}
  & \Rightarrow 4\cos x\left( {{\sin }^{2}}2x \right) \\
 & \left[ \sin 2x=2\sin x\cos x \right] \\
 & \Rightarrow 4\cos x{{\left( 2\sin x\cos x \right)}^{2}} \\
 & \Rightarrow 4\cos x{{\left( 2\sin x\cos x \right)}^{2}} \\
 & \Rightarrow 4\cos x\left( 4{{\sin }^{2}}x{{\cos }^{2}}x \right) \\
 & \Rightarrow 16{{\sin }^{2}}x{{\cos }^{3}}x \\
\end{align}\]
Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved.

NOTE: Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof.