Question

Show that sin^2x+sin^2(2pie/3+x)+sin^2(2pie/3-x)=3/2

Answer 1

Answer:

Step-by-step explanation:

Sin²x + Sin²(2π/3 + x) + Sin²(2π/3 - x) = 3/2

L.H.S

We know that Sin(A+B) = SinACosB + CosASinB

                       Sin(A-B) = SinACosB - CosASinB

Sin²x + Sin²(2π/3 + x) + Sin²(2π/3 - x)

= Sin²x + [Sin(2π/3).Cosx + Cos(2π/3)Sinx]² + [Sin(2π/3).Cosx - Cos(2π/3)Sinx]²

= Sin²x + [√3/2*Cosx + (-1/2)Sinx]² + [√3/2*Cosx - (-1/2)Sinx]²

= Sin²x + [√3/2Cosx - 1/2Sinx]² + [√3/2Cosx + 1/2Sinx]²

= = Sin²x + 3/4Cos²x + 1/4Sin²x - 2(√3/2Cosx)(1/2Sinx) + 3/4Cos²x + 1/4Sin²x + 2(√3/2Cosx)(1/2Sinx)

= Sin²x + 3/2Cos²x + 1/2Sin²x

= 3/2Sin²x + 3/2Cos²x

= 3/2(Sin²x + Cos²x)

= 3/2.

= R.H.S

Hence proved