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Question: Answered & Verified by Expert
If $A+B+C=\frac{\pi}{3}$ then $\sin \left(\frac{\pi-6 A}{6}\right)+\sin \left(\frac{\pi-6 B}{6}\right)+\sin C=$
MathematicsTrigonometric Ratios & IdentitiesTS EAMCETTS EAMCET 2019 (04 May Shift 2)
Options:
  • A $-1+4 \cos \left(\frac{\pi-6 A}{12}\right) \cos \left(\frac{\pi-6 B}{12}\right) \sin \frac{C}{2}$
  • B $4 \sin \left(\frac{\pi+6 A}{12}\right) \sin \left(\frac{\pi+6 B}{12}\right) \cos \frac{C}{2}$
  • C $1-4 \cos \left(\frac{\pi-6 A}{12}\right) \cos \left(\frac{\pi-6 B}{12}\right) \cos \frac{\pi-6 C}{12}$
  • D $4 \cos \left(\frac{\pi-6 A}{12}\right) \cos \left(\frac{\pi-6 B}{12}\right) \sin \frac{C}{2}$
Solution:
1513 Upvotes Verified Answer
The correct answer is: $4 \cos \left(\frac{\pi-6 A}{12}\right) \cos \left(\frac{\pi-6 B}{12}\right) \sin \frac{C}{2}$
Given, $A+B+C=\frac{\pi}{3}$
$$
\begin{aligned}
& \sin \left(\frac{\pi-6 A}{6}\right)+\sin \left(\frac{\pi-6 B}{6}\right)+\sin C \\
& =2 \sin \left(\frac{\pi-6 A+\pi-6 B}{12}\right) \\
& \cos \left(\frac{\pi-6 A-\pi+6 B}{12}\right)+\sin C \\
& =2 \sin \left(\frac{\pi}{6}-\left(\frac{A+B}{2}\right)\right) \cos \left(\frac{A-B}{2}\right)+\sin C \\
& =2 \sin \left(\frac{\pi}{6}-\frac{\pi}{6}+\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cos \frac{C}{2}
\end{aligned}
$$

$$
\begin{aligned}
& =2 \sin \frac{C}{2} \cos \frac{A-B}{2}+2 \sin \frac{C}{2} \cos \frac{C}{2} \\
& =2 \sin \frac{C}{2}\left(\cos \frac{A-B}{2}+\cos \frac{C}{2}\right) \\
& =2 \sin \frac{C}{2}\left(2 \cos \left(\frac{\frac{A-B}{2}+\frac{C}{2}}{2}\right) \cos \left(\frac{\frac{A-B}{2}-\frac{C}{2}}{2}\right)\right) \\
& =4 \sin \frac{C}{2} \cos \left(\frac{A+C-B}{4}\right) \cos \left(\frac{A-(B+C)}{4}\right) \\
& =4 \sin \frac{C}{2} \cos \left(\frac{\pi-6 B}{12}\right) \cos \left(\frac{\pi-6 A}{12}\right) \\
& =4 \cos \left(\frac{\pi-6 A}{12}\right) \cos \left(\frac{\pi-6 B}{12}\right) \sin \frac{C}{2}
\end{aligned}
$$

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