prove of sinA*sinA + sin(A+60)sin(A+60) +sin(A-60)sin(A-60)=3/2?

Trigonometric identities used:
• sin(α ± β) = sinα cosβ ± cosα sinβ
• sin²α + cos²α = 1

L.H.S.
= sinA sinA + sin(A+60°) sin(A+60°) +sin(A-60°) sin(A-60°)
= sin²A + sin²(A+60°) +sin²(A-60°)
= sin²A + [sinA cos60° + cosA sin60°]² + [sinA cos60° - cosA sin60°]²
= sin²A + [(1/2)sinA + (√3/2)cosA]² + [(1/2)sinA - (√3/2)cosA]²
= sin²A + [(1/4)sin²A + (√3/2)sinA cosA + (3/4)cos²A] + [(1/4)sin²A - (√3/2)sinA cosA + (3/4)cos²A]
= sin²A + (1/4)sin²A + (√3/2)sinA cosA + (3/4)cos²A + (1/4)sin²A - (√3/2)sinA cosA + (3/4)cos²A
= (3/2)sin²A + (3/2)cos²A
= (3/2)(sin²A + cos²A)
= 3/2
= R.H.S.

Hence, sinA sinA + sin(A+60°) sin(A+60°) +sin(A-60°) sin(A-60°) = 3/2

Do you know this identity?

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: b = 60

sin(a + 60) = sin(a).cos(60) + cos(a).sin(60)

sin(a + 60) = sin(a).(1/2) + cos(a).[(√3)/2]

sin(a + 60) = (1/2).[sin(a) + (√3).cos(a)] ← memorize this result as (1)



Do you know this identity?

sin(a - b) = sin(a).cos(b) - cos(a).sin(b) → suppose that: b = 60

sin(a - 60) = sin(a).cos(60) - cos(a).sin(60)

sin(a - 60) = sin(a).(1/2) - cos(a).[(√3)/2]

sin(a - 60) = (1/2).[sin(a) - (√3).cos(a)] ← memorize this result as (2)


Your expression

= sin(a).sin(a) + sin(a + 60).sin(a + 60) + sin(a - 60).sin(a - 60)

= sin²(a) + sin²(a + 60) + sin²(a - 60) → recall (1): sin(a + 60) = (1/2).[sin(a) + (√3).cos(a)]

= sin²(a) + { (1/2).[sin(a) + (√3).cos(a)] }² + sin²(a - 60) → recall (2): sin(a - 60) = (1/2).[sin(a) - (√3).cos(a)]

= sin²(a) + { (1/2).[sin(a) + (√3).cos(a)] }² + { (1/2).[sin(a) - (√3).cos(a)] }²

= sin²(a) + (1/4).[sin(a) + (√3).cos(a)]² + (1/4).[sin(a) - (√3).cos(a)]²

= sin²(a) + (1/4).[sin²(a) + (2√3).sin(a).cos(a) + 3.cos²(a)] + (1/4).[sin²(a) - (2√3).sin(a).cos(a) + 3.cos²(a)]

= sin²(a) + (1/4).[sin²(a) + (2√3).sin(a).cos(a) + 3.cos²(a) + sin²(a) - (2√3).sin(a).cos(a) + 3.cos²(a)]

= sin²(a) + (1/4).[sin²(a) + 3.cos²(a) + sin²(a) + 3.cos²(a)]

= sin²(a) + (1/4).[2.sin²(a) + 6.cos²(a)]

= sin²(a) + (1/2).sin²(a) + (3/2).cos²(a)

= (3/2).sin²(a) + (3/2).cos²(a)

= (3/2).[sin²(a) + cos²(a)] → recall the famous formula: sin²(a) + cos²(a) = 1

= (3/2).[1]

= 3/2

sin A sin A = sin^2 A = (1- cos 2A)/2

sin A sin B = (1/2) [ cos(A-B) - cos(A+B)]
sin A sin A = (1/2) [ cos(A-A) - cos(A+A)] = (1-cos A)/2

sin(A+60)sin(A+60) = (1-cos(2A+120)) )/2
sin(A-60)sin(A-60) = (1- cos((2A-120)) )/2

= (1-cos 2A)/2 + (1-cos(2A+120)/2 +(1-cos(2A-120))/2
= (3 -cos(2A)-cos(2A+120)-cos(2A-120)) /2 = 3/2
if (-cos(2A)-cos(2A+120)-cos(2A-120)) /2 = 0

We need to prove (-cos(2A)-cos(2A+120)-cos(2A-120)) /2 = 0
or
(-cos(2A)-cos(2A+120)-cos(2A-120)) = 0

cos(120)=-cos(60)=-1/2
cos(2A+120) = cos(2A)cos(120)-sin(2A)sin(120) = -cos(2A)/2-sin(2A)sin(120)
cos(2A-120) = cos(2A)cos(120)+sin(A)sin(120) = -cos(2A)/2 + sin(2A)sin(120)

-cos(2A)-cos(2A+120)-cos(2A-120) = -cos(2A) -[-cos(2A)/2-sin(2A)sin(120)] - [-cos(2A)/2 + sin(2A)sin(120)]
= -cos(2A) +cos(2A)/2 +sin(2A)sin(120) + cos(2A)/2 -sin(A)sin(120)
= 0

Therefore:
(-cos(2A)-cos(2A+120)-cos(2A-120)) /2 = 0
or
sinA*sinA + sin(A+60)sin(A+60) +sin(A-60)sin(A-60)=3/2