If $ \cos (\alpha+\beta)=0 $, then $ \sin (\alpha-\beta) $ can be reduced to
(A) $ \cos \beta $
(B) $ \cos 2 \beta $
(C) $ \sin \alpha $
(D) $ \sin 2 \alpha $
Given:
\( \cos (\alpha+\beta)=0 \)
To do:
We have to find the value of \( \sin (\alpha-\beta) \)
Solution:
$\cos (\alpha+\beta) =0$
$=\cos 90^{\circ}$ [Since $\cos 90^{\circ}=0$]
This implies,
$\alpha+\beta =90^{\circ}$
$\alpha =90^{\circ}-\beta$
$\sin\ (\alpha-\beta) =\sin (90^{\circ}-\beta-\beta)$
$=\sin\ (90^{\circ}-2 \beta)$
$=\cos 2 \beta$ [Since $\sin\ (90^{\circ}-\theta)=\cos \theta$]
Hence, $\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.
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