Question

Prove the following:
$\dfrac{\sin x-\sin y}{\cos x+\cos y}=\tan \dfrac{x-y}{2}$

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 $\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ and $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ 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:
$\dfrac{\sin x-\sin y}{\cos x+\cos y}=\tan \dfrac{x-y}{2}$
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:
$\begin{aligned} & \sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)................\left( 1 \right) \\\ & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)...............\left( 2 \right) \\\ & \dfrac{\sin \theta }{\cos \theta }=\tan \theta ............................................................\left( 3 \right) \\\ \end{aligned}$
Now, we will use the above five formulas to simplify the term on the left-hand side.
On the left-hand side, we have $\dfrac{\sin x-\sin y}{\cos x+\cos y}$ . Then,
Now, we will use the formula from the equation (1) to write $\sin x-\sin y=2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)$ in the term on the left-hand side. Then,
$\begin{aligned} & \dfrac{\sin x-\sin y}{\cos x+\cos y} \\\ & \Rightarrow \dfrac{2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)}{\cos x+\cos y} \\\ \end{aligned}$
Now, we will use the formula from the equation (2) to write $\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)$ in the above expression. Then,
$\begin{aligned} & \dfrac{2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)}{\cos x+\cos y} \\\ & \Rightarrow \dfrac{2\cos \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{x-y}{2} \right)}{2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)} \\\ & \Rightarrow \dfrac{\sin \left( \dfrac{x-y}{2} \right)}{\cos \left( \dfrac{x-y}{2} \right)} \\\ \end{aligned}$
Now, we will use the formula from the equation (3) to write $\dfrac{\sin \left( \dfrac{x-y}{2} \right)}{\cos \left( \dfrac{x-y}{2} \right)}=\tan \left( \dfrac{x-y}{2} \right)$ in the above expression. Then,
$\begin{aligned} & \dfrac{\sin \left( \dfrac{x-y}{2} \right)}{\cos \left( \dfrac{x-y}{2} \right)} \\\ & \Rightarrow \tan \left( \dfrac{x-y}{2} \right) \\\ \end{aligned}$
Now, from the above result, we conclude that the value of the expression $\dfrac{\sin x-\sin y}{\cos x+\cos y}$ will be equal to the value of the expression $\tan \dfrac{x-y}{2}$ . Then,
$\dfrac{\sin x-\sin y}{\cos x+\cos y}=\tan \dfrac{x-y}{2}$
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, $\dfrac{\sin x-\sin y}{\cos x+\cos y}=\tan \dfrac{x-y}{2}$ .
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 $\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ and $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.