Question
Class 11MathematicsTrigonometric Functions

a cosA + b cosB + c cosC = 2a sinB sinC

Verified Answer

L.H.S. = a cos A + b cos B + c cos C 

= k sin A cos A + k sin B cos B + k sin C cosC

Multiply and divide by 2, = (k/2) [sin 2A + sin 2B + sin 2C] 

= (k/2) [2sin (A + B) cos (A - B) + 2 sin C cos C] 

= (k/2) × 2 [sin (π - C) cos (A - B) + sin C cos C] 

= k [sin C cos (A - B) + sin C cos C] 

= k sin C [cos (A - B) + cos C] = 

k sin C [2 cos ((A - B + C)/2) cos ((A - B - C)/2)] 

= k sin C [2 cos ((π - 2B)/2) cos ((2A - π)/2)] 

= 2k sinC sinB sinA = 2a sinB sinC = R.H.S.