Question

sin²x+sin²(x+π/3)+sin²(x-π/3)=3/2​

Answer 1

To prove :

sin²x + sin²(x + π/3) + sin²(x - π/3) = 3/2​

• Starting with left hand side :

sin² x + sin²(x + π/3) + sin²(x - π/3)

= sin²x + { sin (x + π/3) }² + { sin (x - π/3) }²

= sin²x + ( sin x.cos π/3 + cos x.sin π/3 )² + ( sin x.cos π/3 - cos x.sin π/3 )²

[ sin (a + b) = sin a.cos b + cos a.sin b

 sin ( a - b) = sin a.cos b - cos a.sin b ]

=  sin²x + { sin x. (1 / 2) + cos x.( √3 / 2) }² + { sin x. (1 / 2) - cos x.( √3 / 2) }²

[ cos π/3 = cos 60° = 1 / 2

  sin π/3 = sin 60° = √3 / 2 ]

= sin²x + { (sin x) / 2 + (√3 cos x) / 2 }² + { (sin x) / 2 - (√3cos x) / 2  }²

= sin²x + { (sin x + √3 cos x) / 2 }² + { (sin x - √3 cos x) / 2 }²

= sin²x + (sin x + √3 cos x)² / 2² + (sin x - √3 cos x)² / 2²

= sin²x + (sin x + √3 cos x)² / 4 + (sin x - √3 cos x)² / 4

= sin²x + { (sin x + √3 cos x)² + (sin x - √3 cos x)² } / 4

= sin²x + 2 { (sin x)²  + (√3 cos x)² } / 4

[ (a + b)² - (a - b)² = 2 (a² + b²), where a = sin x, and b = √3 cos x ]

= sin²x + (sin²x + 3 cos²x) / 2

[ (√3)² = √3 × √3 = 3 ]

= (2 sin²x + sin²x + 3 cos²x) / 2

= ( 3 sin²x + 3 cos²x ) / 2

= 3 (sin²x + cos²x) / 2

= (3.1) / 2

[ sin²x + cos²x = 1 ]

= 3 / 2  

= R.H.S.

Hence, proved.