Question

write the expression as a sum or difference of trigonometric functions.
6 sin 2x sin 3x

6 sin 2x sin 3x = \square

Explanation:

Step1: Recall Product-to-Sum Formula

The product - to - sum formula for \(\sin A\sin B\) is \(\sin A\sin B=\frac{1}{2}[\cos(A - B)-\cos(A + B)]\)

Step2: Apply the Formula to \(6\sin2x\sin3x\)

First, we know that for \(A = 2x\) and \(B=3x\), \(\sin2x\sin3x=\frac{1}{2}[\cos(2x - 3x)-\cos(2x + 3x)]\)
Simplify the arguments of the cosine functions:
\(\cos(2x-3x)=\cos(-x)=\cos x\) (since \(\cos(-\theta)=\cos\theta\)) and \(\cos(2x + 3x)=\cos(5x)\)
So, \(\sin2x\sin3x=\frac{1}{2}[\cos x-\cos(5x)]\)

Then, multiply both sides by 6:
\(6\sin2x\sin3x=6\times\frac{1}{2}[\cos x-\cos(5x)]\)
\(6\times\frac{1}{2}=3\), so \(6\sin2x\sin3x = 3\cos x-3\cos(5x)\)

Answer:

\(3\cos x - 3\cos(5x)\)