Question

sin A+ sin (120+A) + sin(240+A)=​

Answer 1

To prove that->

sinA + sin (120′+A) + sin (240′+A) = 0————→ (1)

Observe the L.H.S of equation (1),

sin A + sin (120+A) + sin (240+A) = sin A + (sin 120.cos A + cos 120.sinA )+ (sin 240.cosA + cos 240.sin A )

(∵ sin(A+B) = sinA.cosB +cosA.sinB)

sinA + sin (120+A) + sin (240+A) = sin A + √3/2 . cosA+ (-1/2) . sin A +

(-√3/2) . cosA + (-1/2) . sin A

(∵ sin120′ = √3/2, sin240′ = -√3/2) , (cos 120′ = cos 240′ = -1/2)

sinA+ sin (120+A) + sin (240+A) = sin A + √3/2 . cosA - 1/2 . sinA

-√3/2 . cosA - 1/2 . sinA

sinA + sin (120+A) + sin (240+A) = sinA - [(1/2 + 1/2) . sin A]

sinA + sin (120+A) + sin (240+A) = sinA - (1) . sinA

sinA + sin (120+A) + sin (240+A) = 0

This is equal to R.H.S of equation (1).

Hence proved.