Question

Prove the following:
$\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x$

Answer 1

Hint:To prove the above expression, solve the L.H.S of the expression. In the numerator of the L.H.S, use the identity for $\cos C+\cos D$ in $\cos 4x+\cos 2x$ and in the denominator of L.H.S, use the identity $\sin C+\sin D$ in $\sin 4x+\sin 2x$. Then simplify the expression.

Complete step-by-step answer:
The expression given in the question is:
$\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}=\cot 3x$
Now, we are going to simplify the L.H.S of the above equation.
$\dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x}$
The numerator of the above expression is in the form of $\cos C+\cos D$ so we can apply the identity of $\cos C+\cos D$ in $\cos 4x+\cos 2x$.
$\begin{align}
  & \cos 4x+\cos 2x=2\cos \left( \dfrac{4x+2x}{2} \right)\cos \left( \dfrac{4x-2x}{2} \right) \\
 & \Rightarrow \cos 4x+\cos 2x=2\cos \left( 3x \right)\cos \left( x \right) \\
\end{align}$
If you can see the denominator of the expression, $\sin 4x+\sin 2x$ is in the form of $\sin C+\sin D$ so we can apply the identity of $\sin C+\sin D$ in the expression $\sin 4x+\sin 2x$.
$\begin{align}
  & \sin 4x+\sin 2x=2\sin \left( \dfrac{4x+2x}{2} \right)\cos \left( \dfrac{4x-2x}{2} \right) \\
 & \Rightarrow \sin 4x+\sin 2x=2\sin \left( 3x \right)\cos \left( x \right) \\
\end{align}$
Substituting the value of expressions $\sin 4x+\sin 2x\And \cos 4x+\cos 2x$ in the L.H.S expression given in the question we get,
$\begin{align}
  & \dfrac{\cos 4x+\cos 3x+\cos 2x}{\sin 4x+\sin 3x+\sin 2x} \\
 & =\dfrac{2\cos 3x\cos x+\cos 3x}{2\sin 3x\cos x+\sin 3x} \\
 & =\dfrac{\cos 3x(2\cos x+1)}{\sin 3x(2\cos x+1)} \\
\end{align}$
In the above expression $2\cos x+1$ will be cancelled out from the numerator and the denominator.
$\begin{align}
  & \dfrac{\cos 3x}{\sin 3x} \\
 & =\cot 3x \\
\end{align}$
From the above calculation, the result of L.H.S in the given expression has come out to be $\cot 3x$ which is equal to R.H.S of the given expression.
Hence, we have proved L.H.S = R.H.S of the given expression.

Note: You might be wondering why we have applied $\cos C+\cos D$ for $\cos 4x+\cos 2x$ in the numerator and $\sin C+\sin D$ for $\sin 4x+\sin 2x$ in the denominator because when we take$\cos 4x+\cos 2x$ then after applying the identity then expression converts to $\cos 3x$. Similarly, in the denominator we have got $\sin 3x$ and the R.H.S of the given expression contains $\cot 3x$. So, to equate both the sides, we have selectively taken the angles $4x$ and $2x$ of cosine and sine in the numerator and denominator respectively.