sin 2A + sin 2B + sin (A+B) = 4 cos A cos B sin(A+B)
Answers
Answer:
To prove---> There is a mistake in
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question i think it is like this
Sin2A + Sin2B +Sin2(A+B)=4CosA CosB
Sin(A+B)
Proof---> LHS=
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Sin 2A + Sin 2B +Sin 2(A + B)
We have a formula
Sinx + Siny= 2 Sin(x+y/2) Cos (x-y/2)
applying it here
= 2 Sin (2A + 2B/2) Cos (2A - 2B/2)
+Sin2(A+B)
We have a formula
Sin 2θ = 2 Sinθ Cosθ
Applying it here
= 2 Sin(A + B) Cos (A - B) + 2Sin (A+B)
Cos(A+B)
= 2 Sin (A + B){Cos(A-B) + Cos (A+B)}
We have a formula
cosx + Cosy = 2 Cos(x+y/2) Cos(x-y/2)
Here
x= A+B y= A-B
x+y = A+B + A-B
= 2A
x-y = A+B -(A-B)
= A+B -A+B
= 2B
Now returning to original problem
= 2 Sin(A+B) {2 Cos (2A/2) Cos (2B/2)}
= 2 Sin(A+B) (2 Cos A Cos B )
= 4 CosA Cos B Sin (A + B) =RHS