Question

If cos(α – β) + cos(β – γ) + cos(γ – α) = `- 3/2`, then prove that cos α + cos β + cos γ = sin α + sin β + sin γ = 0

Answer 1

⇒ Given cos(α – β) + cos(β – γ) + 2cos(γ – α) = `- 3/2`

2 cos(α – β) + 2cos(β – γ) + 2cos(γ – α) = – 3

2cos(α – β) + 2cos(β – γ) + 2cos(γ – α) + 3 = 0

[2 cos α cos β + 2 sin α sin β] + [2 cos β cos γ + 2 sin β sin γ] + [2 cos γ cos α + sin γ sin α] + 3 = 0

= [2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α] + [2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α] + (sin² α + cos² α) + (sin² β + cos² β) + (sin² γ + cos² γ) = 0

⇒ (cos² α + cos² β + cos² γ + 2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (sin² α + sin² β) + (sin² γ + 2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α) = 0

(cos α + cos β + cos γ)² + (sin α + sin β + sin γ)² = 0

= (cos α + cos β + cos γ) = 0 and sin α + sin β + sin γ = 0

Hence proved