(i) Consider sin (A – B) sin C
= (sin A cos B – cos A sin B) sin C
= sin A cos B sin C – cos A sin B sin C … (1)
Similarly sin(B – C) sin A = sin B cos C sin A – cos B sin C sin A … (2)
[Replace A by B, B by C, C by A in (1)] and sin(C – A) sin B [Replace A by B, B by C, C by A in (2)]
= sin C cos A sin B – cos C sin A sin B … (3)
Adding (1), (2) and (3) we get
sin (A – B) sin C + sin (B – C) sin A + sin(C – A) sin B = 0
(ii) 2 cos\(\frac{\pi}{13}\) cos\(\frac{9\pi}{13}\) + cos\(\frac{3\pi}{13}\) + cos\(\frac{5\pi}{13}\) = 0
LHS = 2 cos\(\frac{\pi}{13}\) cos\(\frac{9\pi}{13}\) + cos\(\frac{3\pi}{13}\) + cos\(\frac{5\pi}{13}\)
[∵ cos C + cos D = 2 cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))]
[∵ cos(-θ) = cos θ]
[take 2 cos \(\frac{\pi}{3}\) as commom]
Hence proved.
