Question

if sin alpha + sin beta=a and cos alpha+ cos beta = b, then prove that 1) sin(alpha+beta)=2ab/a^2 + b^2.

Answer 1

Understanding:
1) sinα + sinβ = 2sin((α+β)/2) cos((α-β)/2)
2) cosα+cosβ = 2cos((α+β)/2)cos((α-β)/2)

Steps:
1) Convert a and b into sum and difference of angles by using above identity.
2)Mulitply a with b.
3) Add Square a with square b.
4)Divide 2) and 3).

Answer 2

Answer:

  $\frac{2ab}{a^{2}+b^{2} }$ = sin(α+β)

Step-by-step explanation:

We have to prove that sin(α+β) = $\frac{2ab}{a^{2}+b^{2} }$

'a' = sinα + sinβ

  a = 2 sin($\frac{\alpha +\beta }{2}$) cos($\frac{\alpha -\beta }{2}$)

'b' = cosα+cosβ

  b    = 2 cos($\frac{\alpha +\beta }{2}$) cos($\frac{\alpha -\beta }{2}$)

'a b' =  4 sin($\frac{\alpha +\beta }{2}$) cos($\frac{\alpha +\beta }{2}$) cos($\frac{\alpha -\beta }{2}$)

ab    = 2 sin($\frac{\alpha +\beta }{2}$) cos²($\frac{\alpha -\beta }{2}$)

a² + b²= 4 cos²($\frac{\alpha -\beta }{2}$)

∴ $\frac{2ab}{a^{2}+b^{2} }$ = sin(α+β)

Hence proved.

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