Question

Prove that sin^{​​​​2}​​​ a + sin ² ​​​​​​(120°- a) + sin ²(120° - a) = 3/2

Answer 1

Now, cos 2A = 1 - 2 sin^2 A ==> sin^2 A = (1 - cos 2A) / 2

==> LHS = [ 1 - cos 2(120-A)]/2 + [1 - cos 2A]/2 + [ 1 - cos 2(120+A)] / 2

= (1/2) * [ 3 - [ cos 2(120-A) + cos 2A + cos 2(120+A) ] ]

Now, cos A + cos B = 2 cos ½ (A + B) cos ½ (A − B)

==> LHS = (1/2) * [ 3 - [ cos 2A + [ cos 2(120-A) + cos 2(120+A) ] ] ]

= (1/2) * [ 3 - [ cos 2A +

2 cos ( (½) * 2(120-A+ 120+A)) cos ( (½) * 2(120-A- 120-A)) ]

= (1/2) * [ 3 - [ cos 2A + 2 cos 240 cos ( -2A) ]

= (1/2) * [ 3 - [ cos 2A + 2 (-1/2) cos (2A) ]

= (1/2) * [ 3 - [ cos 2A - cos (2A) ] ]

= 3/2