Question

Given that: (1 + cosα) (1 + Cosβ) (1 + cosγ) = (1 - cosα)(1 – cosβ) (1 – cos γ) .Show that one of the values of each member of this equality is sinα sinβ sinγ.



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Answer 1

Answer:

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

Answer 2

$\huge\mathfrak\pink{Answer}$

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