Question

Applying the product-to-sum identity to \(\sin 2x\cos 3x\) gives which expression?

Options

• \(\frac{1}{2}[\sin 5x-\sin x]\)

• \(\sin 5x-\sin x\)

• \(\frac{1}{2}[\sin 5x+\sin x]\)

• \(\frac{1}{2}[\cos 5x-\cos x]\)

Answer 1

Set \(A=2x\) and \(B=3x\) in \(\sin A\cos B=\frac{1}{2}[\sin(A+B)+\sin(A-B)]\). This gives \(\frac{1}{2}[\sin 5x+\sin(-x)] = \frac{1}{2}[\sin 5x-\sin x]\).