Question

If we have given $\sin \theta = \sin \alpha $
Then find which option satisfies the given identity.
$
  {\text{A}}{\text{. }}\dfrac{{\theta + \alpha }}{2}{\text{ is any odd multiple of }}\dfrac{\pi }{2}{\text{ and }}\dfrac{{\theta - \alpha }}{2}{\text{ is any multiple of }}\pi \\
  {\text{B}}{\text{. }}\dfrac{{\theta + \alpha }}{2}{\text{ is any even multiple of }}\dfrac{\pi }{2}{\text{ and }}\dfrac{{\theta - \alpha }}{2}{\text{ is any multiple of }}\pi \\
  {\text{C}}{\text{. }}\dfrac{{\theta + \alpha }}{2}{\text{ is any multiple of }}\dfrac{\pi }{2}{\text{ and }}\dfrac{{\theta - \alpha }}{2}{\text{ is any odd multiple of }}\pi \\
  {\text{D}}{\text{. }}\dfrac{{\theta + \alpha }}{2}{\text{ is any multiple of }}\dfrac{\pi }{2}{\text{ and }}\dfrac{{\theta - \alpha }}{2}{\text{ is any even multiple of }}\pi \\
$

Answer 1

- Hint:- In this question first we have to convert the given question ($\sin \theta = \sin \alpha $) into a form in which we can equate terms involved in it to zero, using trigonometric identities. Then, use the condition $\sin (a)$ and $\cos (b)$ are zero to get the result in required form.

Complete step-by-step solution -

Given: $\sin \theta = \sin \alpha $ eq.1
$ \Rightarrow \sin \theta - \sin \alpha {\text{ = 0}}$

We know, from trigonometric identities
$\sin (a) - \sin (b) = 2\cos \left( {\dfrac{{a + b}}{2}} \right).\sin \left( {\dfrac{{a - b}}{2}} \right)$
On applying above formula on eq.1 we get
$
   \Rightarrow \sin \theta - \sin \alpha {\text{ = 0}} \\
   \Rightarrow 2\cos \left( {\dfrac{{\theta + \alpha }}{2}} \right).\sin \left( {\dfrac{{\theta - \alpha }}{2}} \right) = 0{\text{ eq}}{\text{.2}} \\
$
We know,
If $\sin \theta = 0$
Then, $\theta = n\pi $ eq.3
\[{\text{where }}n = {\text{integer}}\]
And if $\cos \theta = 0$
Then, $\theta = \dfrac{{(2n + 1)\pi }}{2}$ eq.4
\[{\text{where }}n = {\text{integer}}\]
Now, using eq.3 and eq.4 we can write eq.2 as
$
   \Rightarrow 2\cos \left( {\dfrac{{\theta + \alpha }}{2}} \right).\sin \left( {\dfrac{{\theta - \alpha }}{2}} \right) = 0 \\
   \Rightarrow \cos \left( {\dfrac{{\theta + \alpha }}{2}} \right) = 0{\text{ or }}\sin \left( {\dfrac{{\theta - \alpha }}{2}} \right) = 0 \\
$
On solving these two conditions separately we get
$
   \Rightarrow \cos \left( {\dfrac{{\theta + \alpha }}{2}} \right) = 0{\text{ }} \\
   \Rightarrow \left( {\dfrac{{\theta + \alpha }}{2}} \right) = \dfrac{{(2n + 1)\pi }}{2}{\text{ \{ from eq}}{\text{.4\} }} \\
  {\text{where }}n = {\text{integer}} \\
$
Therefore, \[\left( {\dfrac{{\theta + \alpha }}{2}} \right)\] is the odd multiple of $\dfrac{\pi }{2}$. Statement 1
Now, consider \[\sin \left( {\dfrac{{\theta - \alpha }}{2}} \right) = 0\]
$
   \Rightarrow \sin \left( {\dfrac{{\theta - \alpha }}{2}} \right) = 0 \\
   \Rightarrow \left( {\dfrac{{\theta - \alpha }}{2}} \right) = n\pi {\text{ \{ from eq}}{\text{.3\} }} \\
  {\text{where }}n = {\text{integer}} \\
 $
Therefore, $\left( {\dfrac{{\theta - \alpha }}{2}} \right)$ is any multiple of $\pi $. Statement 2
Hence, from Statement 1 and Statement 2, option 1 is correct.

Note: - Whenever you get this type of question the key concept to solve this is to learn all the trigonometric identities. And also you must have to learn the various conditions of trigonometric angles like $\sin \theta ,\cos \theta $ etc. when equating them to zero. Remember $\sin (a) - \sin (b) = 2\cos \left( {\dfrac{{a + b}}{2}} \right).\sin \left( {\dfrac{{a - b}}{2}} \right)$ these formulas for easy solving.