Question

Prove the following:
$\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}$

Answer 1

Hint: For solving this question, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use trigonometric formulas like $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ and $\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Complete step-by-step answer:

Given:
We have to prove the following equation:
$\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & \cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)...................\left( 1 \right) \\
 & \sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)......................\left( 2 \right) \\
\end{align}$
Now, we will use the above two formulas to simplify the term on the left-hand side.
On the left-hand side, we have $\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}$ .
Now, we will use the formula from the equation (1) to write $\cos 9x-\cos 5x=-2\sin 7x\sin 2x$ in the term on the left-hand side. Then,
$\begin{align}
  & \dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x} \\
 & \Rightarrow \dfrac{-2\sin \left( \dfrac{9x+5x}{2} \right)\sin \left( \dfrac{9x-5x}{2} \right)}{\sin 17x-\sin 3x} \\
 & \Rightarrow \dfrac{-2\sin 7x\sin 2x}{\sin 17x-\sin 3x} \\
\end{align}$
Now, we will use the formula from the equation (2) to write $\sin 17x-\sin 3x=2\cos 10x\sin 7x$ in the above expression. Then,
$\begin{align}
  & \dfrac{-2\sin 7x\sin 2x}{\sin 17x-\sin 3x} \\
 & \Rightarrow \dfrac{-2\sin 7x\sin 2x}{2\cos \left( \dfrac{17x+3x}{2} \right)\sin \left( \dfrac{17x-3x}{2} \right)} \\
 & \Rightarrow \dfrac{-2\sin 7x\sin 2x}{2\cos 10x\sin 7x} \\
 & \Rightarrow \dfrac{-\sin 7x\sin 2x}{\sin 7x\cos 10x} \\
 & \Rightarrow -\dfrac{\sin 2x}{\cos 10x} \\
\end{align}$
Now, from the above result, we conclude that the value of the expression $\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}$ will be equal to the value of the expression $-\dfrac{\sin 2x}{\cos 10x}$ . Then,
$\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}$
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, $\dfrac{\cos 9x-\cos 5x}{\sin 17x-\sin 3x}=-\dfrac{\sin 2x}{\cos 10x}$ .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question. After that, we should proceed in a stepwise manner and apply trigonometric formulas like $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ and $\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.