Question -
Answer -
(i) sin α + sin β + sin γ – sin (α + β + γ) = 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2
Let us consider LHS:
sin α + sin β + sin γ – sin (α + β + γ)
By using the formulas,
Sin A + sin B = 2 sin (A+B)/2 cos (A-B)/2
Sin A – sin B = 2 cos (A+B)/2 sin (A-B)/2
= RHS
Hence proved.
(ii) cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) = 4 cos A cos B cos C
Let us consider LHS:
cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C)
so,
cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) =
={cos (A + B + C) + cos (A – B + C)} + {cos (A + B – C) + cos (-A + B + C)}
By using the formula,
Cos A + cos B = 2 cos (A+B)/2 cos (A-B)/2
= 4 cos A cos B cos C
= RHS
Hence proved.