Given that:
(1+cosα)(1+cosβ)(1+cosγ)=(1-cosα)(1-cosβ)(1-cosγ)
Show that one of the values of each member of this equality is sinα sinβ sinγ
Answers
sinα sinβ sinγ is one of the values of each member of this equality
Step-by-step explanation:
(1+cosα)(1+cosβ)(1+cosγ)=(1-cosα)(1-cosβ)(1-cosγ)
LHS = (1+cosα)(1+cosβ)(1+cosγ)
= 2Cos²(α/2)2Cos²(β/2)2Cos²(γ/2)
= 8Cos²(α/2)Cos²(β/2)Cos²(γ/2)sinα sinβ sinγ/sinα sinβ sinγ
= 2Cos²(α/2)Cos²(β/2)Cos²(γ/2)sinα sinβ sinγ/2sin(α/2)Cos(α/2) 2sin(β/2)Cos(β/2)2sin(γ/2)Cos(γ/2)
= Cot(α/2)Cot(β/2)Cot(γ/2)sinα sinβ sinγ
RHS = (1-cosα)(1-cosβ)(1-cosγ)
= 2Sin²(α/2)2Sin²(β/2)2Sin²(γ/2)
= 8Sin²(α/2)Sin²(β/2)Sin²(γ/2)sinα sinβ sinγ/sinα sinβ sinγ
= 8Sin²(α/2)Sin²(β/2)Sin²(γ/2)sinα sinβ sinγ/2sin(α/2)Cos(α/2) 2sin(β/2)Cos(β/2)2sin(γ/2)Cos(γ/2)
= Tan(α/2)Tan(β/2)Tan(γ/2)sinα sinβ sinγ
Cot(α/2)Cot(β/2)Cot(γ/2)sinα sinβ sinγ = Tan(α/2)Tan(β/2)Tan(γ/2)sinα sinβ sinγ
Hence sinα sinβ sinγ is one of the values of each member of this equality
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