Prove that cos alpha+cos beta + cos gamma + cos (alpha+ beta + gamma) = 4 cos alpha + beta/2 cos beta +gamma/2 cos gamma+alpha/2
Answers
Answered by
99
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)
Answered by
29
Answer:
Hey Mate,
See attachment
Attachments:
Similar questions