Question

Prove the following expression: $\cot 4x\left( \sin 5x+\sin 3x \right)=\cot x\left( \sin 5x-\sin 3x \right)$.

Answer 1

Hint: To solve this question, we have to convert all the terms of any of the sides in terms of either sin or cos. We know that, $\sin a+\sin b=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$ . We also know that, $2\sin c\cos d=\sin \left( c+d \right)+\sin \left( c-d \right)$ and we also know that, $\sin \left( -\theta \right)=-\sin \theta $. By using these relations, we can prove the given expression.

Complete step-by-step answer:
In this question, we have been asked to prove that $\cot 4x\left( \sin 5x+\sin 3x \right)=\cot x\left( \sin 5x-\sin 3x \right)$. Let us consider the left hand side or LHS first, that is, $\cot 4x\left( \sin 5x+\sin 3x \right)$. Now we know that, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$, so by substituting it in the LHS, we can write the LHS as, $\dfrac{\cos 4x}{\sin 4x}\left( \sin 5x+\sin 3x \right)$. We know that, $\sin a+\sin b=2\left[ \sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right) \right]$. So, we get LHS as, $\begin{align}
  & \dfrac{\cos 4x}{\sin 4x}\left[ 2\left[ \sin \left( \dfrac{5x+3x}{2} \right) \right]\left[ \cos \left( \dfrac{5x-3x}{2} \right) \right] \right] \\
 & \Rightarrow \dfrac{\cos 4x}{\sin 4x}\left[ 2\left[ \sin \left( \dfrac{8x}{2} \right) \right]\left[ \cos \left( \dfrac{2x}{2} \right) \right] \right] \\
 & \Rightarrow \dfrac{\cos 4x}{\sin 4x}\left[ 2\left( \sin 4x \right)\left( \cos x \right) \right] \\
\end{align}$
Cancelling the similar terms, we get LHS as,
$2\left( \cos 4x \right)\left( \cos x \right)$
Now, we will multiply the numerator and the denominator by $\sin x$, so we get LHS as, $\begin{align}
  & 2\left( \cos 4x \right)\left( \cos x \right)\dfrac{\left( \sin x \right)}{\left( \sin x \right)} \\
 & \Rightarrow 2\left( \cos 4x \right)\left( \sin x \right)\left( \dfrac{\cos x}{\sin x} \right) \\
\end{align}$
We also know that, $2\sin c\cos d=\sin \left( c+d \right)+\sin \left( c-d \right)$, so by substituting that we get the LHS as, $\left[ \sin \left( 4x+x \right)+\sin \left( x-4x \right) \right]\left[ \dfrac{\cos x}{\sin x} \right]$
On further simplification, we get,
$\begin{align}
  & \left[ \sin 5x+\sin \left( -3x \right) \right]\cot x \\
 & \Rightarrow \cot x\left[ \sin 5x+\sin \left( -3x \right) \right] \\
\end{align}$
We know that $\sin \left( -\theta \right)=-\sin \theta $, so, by substituting the same in the above expression, we get the LHS as, $\cot x\left( \sin 5x-\sin 3x \right)$, which is equal to the right hand side or RHS, which is also, $\cot x\left( \sin 5x-\sin 3x \right)$.
Hence, we have proved the expression given in the question, $\cot 4x\left( \sin 5x+\sin 3x \right)=\cot x\left( \sin 5x-\sin 3x \right)$.

Note: While solving this question, we have to keep in mind what we are trying to prove and then apply the respective formulas accordingly. For example, if we had not multiplied the numerator and the denominator by $\sin x$, then we would have gone in the wrong direction and made the answer longer.