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If angles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are in $\mathrm{AP}$, then what is $\sin A+2 \sin B+\sin C$ equal to?
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Verified Answer
The correct answer is:
$4 \sin B \cos ^{2}\left(\frac{A-C}{4}\right)$
Since, $A, B, C$ are in $A P$. $\therefore B-A=C-B$
$\Rightarrow 2 B=A+C$
But we know $A+B+C=180^{\circ}$
$\Rightarrow \quad 3 B=180^{\circ} \Rightarrow B=60^{\circ}$
Consider $\sin A+2 \sin B+\sin C$
$=2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}+2 \sin B$
$=2 \sin B\left[\cos \frac{A-C}{2}+1\right] \quad(\because A+C=2 B)$
$=2 \sin B\left[2 \cos ^{2}\left(\frac{A-C}{4}\right)\right]$
$=4 \sin B \cos ^{2}\left(\frac{A-C}{4}\right)$
$\Rightarrow 2 B=A+C$
But we know $A+B+C=180^{\circ}$
$\Rightarrow \quad 3 B=180^{\circ} \Rightarrow B=60^{\circ}$
Consider $\sin A+2 \sin B+\sin C$
$=2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}+2 \sin B$
$=2 \sin B\left[\cos \frac{A-C}{2}+1\right] \quad(\because A+C=2 B)$
$=2 \sin B\left[2 \cos ^{2}\left(\frac{A-C}{4}\right)\right]$
$=4 \sin B \cos ^{2}\left(\frac{A-C}{4}\right)$
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