Question

prove that.....
...

.


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Answer 1

Answer:

LHS is solved

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Answer 2

Answer:

Cosα + Cosβ + Cosγ + Cos(α + β+ γ) = 4Cos(α+β/2)Cos(β+γ/2)Cos(α+γ/2)

Step-by-step explanation:

Cosα + Cosβ + Cosγ + Cos(α + β+ γ) = 4Cos(α+β/2)Cos(β+γ/2)Cos(α+γ/2)

Using CosA + CosB =  2Cos((A + B)/2)Cos((A - B)/2)

LHS = Cosα +Cosβ + Cosγ + Cos(α + β+ γ)

= 2Cos((α + β)/2)Cos((α - β)/2)   + 2 Cos((α + β+ 2γ)/2)Cos((-α - β)/2)

Using Cos(-A) = CosA

= 2Cos((α + β)/2)Cos((α - β)/2)   + 2 Cos((α + β+ 2γ)/2)Cos((α + β)/2)

= 2Cos((α + β)/2) (Cos((α - β)/2) + Cos((α + β+ 2γ)/2))

= 2Cos((α + β)/2)  ( 2 Cos((α + γ)/2)Cos((-β - γ)/2)

= 2Cos((α + β)/2)  ( 2 Cos((α + γ)/2) (Cos((β + γ)/2))

= 4 Cos((α + β)/2)Cos((β + γ)/2)Cos((α + γ)/2)

= RHS

QED

Proved

Cosα + Cosβ + Cosγ + Cos(α + β+ γ) = 4Cos(α+β/2)Cos(β+γ/2)Cos(α+γ/2)

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