Question

12. (a) Show that sin A + sin (120° + A) + sin (120° - A) = 0.

Answer 1

$\sin(a) + \cos (30 + a) + \cos(30 - a) \\ \sin(a) + \cos( {30}^{2} - {a}^{2} ) \\ \sin(a) + \cos(900 - {a}^{2} ) \\ \sin(a) - \sin(a) \\ = 0$



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Answer 2

Answer:

Given,

SinA+Sin(120+A)Sin(120-A)=0

Consider L.H.S

=SinA+Sin(120+A)Sin(120-A)

We know that,

[2CosASinB=Sin(A+B)Sin(A-B)]

=SinA+2[Cos(90°+30°)]SinA

=SinA+2(-Sin30°)SinA

because, Cos(90°+theta)=-Sin(theta)

=SinA+2(-1/2)SinA

=SinA-SinA

=0

=R.H.S

So, L.H.S=R.H.S

Hence proved.

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