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If \(\sin \alpha=\sin \beta\) and \(\cos \alpha=\cos \beta\), then \(\alpha-\beta=\) for some integer \(n\).
Options:
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2343 Upvotes
Verified Answer
The correct answer is:
\(2 n \pi\)
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.
\(\Rightarrow 2 \sin \left(\frac{\alpha-\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)=0\) ...(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.
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