If $\sin \alpha=\sin \beta$ and $\cos \alpha=\cos \beta$, then $\alpha-\beta=$ for some integer $n$.

Question

Aicharya · Accepted Answer

It is given that, $\sin \alpha=\sin \beta$ $\Rightarrow 2 \sin \left(\frac{\alpha-\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)=0$ ...(i) and $\cos \alpha=\cos \beta$ $\Rightarrow 2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\beta-\alpha}{2}\right)=0$ ...(ii) from Eqs. (i) and (ii), we get $\begin{aligned} & \sin ^2\left(\frac{\alpha-\beta}{2}\right)\left[\sin ^2\left(\frac{\alpha+\beta}{2}\right)+\cos ^2\left(\frac{\alpha+\beta}{2}\right)\right] =0 \ \Rightarrow & \sin ^2\left(\frac{\alpha-\beta}{2}\right) =0 \ \Rightarrow & \frac{\alpha-\beta}{2} =n \pi \end{aligned}$ $\Rightarrow \alpha-\beta=2 n \pi, n \in$ Integer Hence, option (d) is correct.

If \(\sin \alpha=\sin \beta\) and \(\cos \alpha=\cos \beta\), then \(\alpha-\beta=\) for some integer \(n\).

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