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If $\sin\alpha=\sin\beta$ and $\cos\alpha=\cos\beta$ then

$\begin{array}{1 1}(a)\;\sin\large\frac{\alpha+\beta}{2}\normalsize=0&(b)\;\cos\large\frac{\alpha+\beta}{2}\normalsize=0\\(c)\;\sin\large\frac{\alpha-\beta}{2}\normalsize=0&(d)\;\cos\large\frac{\alpha-\beta}{2}\normalsize=0\end{array}$

1 Answer

$\sin\alpha=\sin\beta$
$\Rightarrow \sin\alpha-\sin\beta=0$
$\Rightarrow 2\cos\large\frac{\alpha+\beta}{2}$$=0$
Or $\sin\large\frac{\alpha-\beta}{2}$$=0$
$\cos\alpha=\cos\beta$
$\Rightarrow \cos\alpha-\cos\beta=0$
$-2\sin\big(\large\frac{\alpha+\beta}{2}\big)$$\sin\big(\large\frac{\alpha-\beta}{2}\big)$$=0$
Either $\sin\big(\large\frac{\alpha+\beta}{2}\big)$$=0$
(Or) $\sin\big(\large\frac{\alpha-\beta}{2}\big)$$=0$
Thus common solution is $\sin\big(\large\frac{\alpha-\beta}{2}\big)$$=0$
answered Oct 8, 2013 by sreemathi.v
 

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