Question

If $A+B+C=\frac{\pi}{3}$ then
$\sin \left(\frac{\pi-6 A}{6}\right)+\sin \left(\frac{\pi-6 B}{6}\right)+\sin C=$

• 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. $4 \cos \left(\frac{\pi-6 A}{12}\right) \cos \left(\frac{\pi-6 B}{12}\right) \sin \frac{C}{2}$

Answer 1

Answer: (D)

Given, $A+B+C=\frac{\pi}{3}$
$\sin \left(\frac{\pi-6 A}{6}\right)+\sin \left(\frac{\pi-6 B}{6}\right)+\sin C$
$=2 \sin \left(\frac{\pi-6 A+\pi-6 B}{12}\right)$
$\cos \left(\frac{\pi-6 A-\pi+6 B}{12}\right)+\sin C$
$=2 \sin \left(\frac{\pi}{6}-\left(\frac{A+B}{2}\right)\right) \cos \left(\frac{A-B}{2}\right)+\sin C$
$=2 \sin \left(\frac{\pi}{6}-\frac{\pi}{6}+\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cos \frac{C}{2}$
$=2 \sin \frac{C}{2} \cos \frac{A-B}{2}+2 \sin \frac{C}{2} \cos \frac{C}{2}$
$=2 \sin \frac{C}{2}\left(\cos \frac{A-B}{2}+\cos \frac{C}{2}\right)$
$=2 \sin \frac{C}{2}\left(2 \cos \left(\frac{\frac{A-B}{2}+\frac{C}{2}}{2}\right) \cos \left(\frac{\frac{A-B}{2}-\frac{C}{2}}{2}\right)\right)$
$=4 \sin \frac{C}{2} \cos \left(\frac{A+C-B}{4}\right) \cos \left(\frac{A-(B+C)}{4}\right)$
$=4 \sin \frac{C}{2} \cos \left(\frac{\pi-6 B}{12}\right) \cos \left(\frac{\pi-6 A}{12}\right)$
$=4 \cos \left(\frac{\pi-6 A}{12}\right) \cos \left(\frac{\pi-6 B}{12}\right) \sin \frac{C}{2}$