Question

If $\sin \left(\theta \right)=\sin \left(\alpha \right)$, then

• A. $\frac{\theta +\alpha }{2}$ is any odd multiple of $\frac{\pi }{2}$ and $\frac{\theta –\alpha }{2}$ is any multiple of $\pi$
• B. $\frac{\theta +\alpha }{2}$ is any odd multiple of $\pi$ and $\frac{\theta –\alpha }{2}$ is any multiple of $\pi$
• C. $\frac{\theta +\alpha }{2}$ is any multiple of $\frac{\pi }{2}$ and $\frac{\theta –\alpha }{2}$ is any even multiple of $\pi$
• D. $\frac{\theta +\alpha }{2}$ is any even multiple of $\frac{\pi }{2}$ and $\frac{\theta –\alpha }{2}$ is any odd multiple of $\pi$

Answer 1

The correct option is A

$\frac{\theta +\alpha }{2}$ is any odd multiple of $\frac{\pi }{2}$ and $\frac{\theta –\alpha }{2}$ is any multiple of $\pi$

Explanation for the correct option

Step 1: Simplification of the given relation

We can write the given relation as

$\sin \left(\theta \right)-\sin \left(\alpha \right)=0$

Use the formula of $\sin \left(A\right)-\sin \left(B\right)$,

$\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{-}\mathbf{sin}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{A}\mathbf{+}\mathbf{B}}{\mathbf{2}}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{A}\mathbf{-}\mathbf{B}}{\mathbf{2}}\mathbf{\right)}$

Let $A=\theta \text{and}B=\alpha$ then we have,

$2\cos \left(\frac{\theta +\alpha }{2}\right)\sin \left(\frac{\theta -\alpha }{2}\right)=0$

From this, we can see that either $\cos \left(\frac{\theta +\alpha }{2}\right)=0$ or $\sin \left(\frac{\theta -\alpha }{2}\right)=0$

Step 2: Check for $\mathbf{cos}\left(\frac{\theta +\alpha }{2}\right)\mathbf{=}\mathbf{0}$

We know that when $\cos \left(\beta \right)=0$ then $\beta$ is equal to $\frac{\pi }{2}$.

Let $n$ be any integer such that when it is multiplied by $\frac{\pi }{2}$ then it gives the value zero when $\cos$ is applied.

If we put the values of $n$ like $0,1,2,3,\dots ,n$ then we can deduce that,

$\cos \left(\frac{n\pi }{2}\right)$ will not be equal to $0$ when $n$ is even so it must be odd. And odd integers are written in the form $2n+1$.

From this, we have that $\left(\frac{\theta +\alpha }{2}\right)$ is any odd multiple of $\frac{\pi }{2}$.

Step 3: Check for $\mathbf{sin}\mathbf{\left(}\frac{\mathbf{A}\mathbf{-}\mathbf{B}}{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{0}$

We know that when $\sin \left(\beta \right)=0$ then $\beta$ is either $0$ or $\pi$

If we multiply any integer $n$ with $\pi$, the value of $\sin \left(n\pi \right)$ will be zero.

From this, we have that $\left(\frac{\theta -\alpha }{2}\right)$ is any multiple of $\pi$.

Hence, the correct option is (A).