MCQ
If $\cos(\alpha+\beta)=0$ then $\sin(\alpha-\beta)=?$
- A$\sin\alpha$
- B$\cos\beta$
- C$\sin2\alpha$
- ✓$\cos2\beta$
✓
Answer
Correct option: D.
$\cos2\beta$
$\cos(\alpha+\beta)=0$
$\Rightarrow\cos(\alpha+\beta)=\cos90^\circ$
$\Rightarrow\alpha+\beta=90^\circ$
$\Rightarrow\alpha=90^\circ-\beta$
Now, $\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)$
$=\sin(90^\circ-2\beta)$
$=\cos2\beta$
$\Rightarrow\cos(\alpha+\beta)=\cos90^\circ$
$\Rightarrow\alpha+\beta=90^\circ$
$\Rightarrow\alpha=90^\circ-\beta$
Now, $\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)$
$=\sin(90^\circ-2\beta)$
$=\cos2\beta$
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