Question

Prove that:

(i) sin α + sin β + sin γ – sin (α + β + γ) = 4 sin (α + β)/2 sin (β + γ)/2 sin (α + γ)/2

(ii) cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C) = 4 cos A cos B cos C

Answer 1

(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 the LHS

sin α + sin β + sin γ – sin (α + β + γ)

On 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

Again on using the formula,

= RHS

Hence proved.

(ii) Given ascos (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 the LHS

cos (A + B + C) + cos (A – B + C) + cos (A + B – C) + cos (-A + B + C)

Therefore,

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)}

On using the formula,

Cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2

Again on using the formula,

= 4 cos A cos B cos C

= RHS

Thus proved.