Question

prove that sinA+sin(120 degree)+sin(240 degree)=0 i need step by step explanation pls answer correctly​

Answer 1

Answer:

How do you prove that SinA+Sin (120'+A) +Sin (240'+A) =0?

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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.

Answer 2

Question:

How do you prove that SinA+Sin (120 degree) +Sin (240 degree) =0?

Prove:

SinA + sin (120 degree) + sin (240 degree) = 0————→ (1)

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

sin A + sin (120°) + sin (240°) = sin A + (sin 120.cos + cos 120.sin )+ (sin 240.cos + cos 240.sin )

(∵ sin( 120°+240°) = sinA.cos +cosA.sin)

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

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

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

=> sin A + sin (120°) + sin (240°) = 0

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

Hence proved...

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