Question

Question number 18 pls help tommorow is my exam

Answer 1

2ab = 2{SinACosA + SinACosB + SinBCosA + SinBCosB}

2ab = 2{Sin(A + B) + SinACosA + SinBCosB}

2ab = 2Sin(A + B) + Sin2A + Sin2B

2ab = 2Sin(A + B) + 2Sin(A + B)Cos(A - B)

2ab = 2Sin(A + B){1 + Cos(A - B)}

a² + b² = 2 + 2(SinASinB + CosACosB)

a² + b² = 2{1 + Cos(A - B)}

⇒ $\frac{2ab}{a^2 + b^2}$ = Sin(A + B)

identities used :

1. SinACosB + CosASinB = Sin(A + B)
2. 2SinACosA = Sin2A
3. SinA + SinB = 2Sin(A + B)/2 × Cos(A + B)/2
4. CosACosB + SinASinB = Cos(A - B)