Question

use a sum - to - product identity to rewrite the expression.
$sin 3alpha+sin 6alpha$
$sin 3alpha+sin 6alpha=square$
(use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Recall the sum - to - product formula

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

Step2: Identify \(A\) and \(B\)

Here \(A = 3\alpha\) and \(B=6\alpha\).

Step3: Calculate \(\frac{A + B}{2}\) and \(\frac{A - B}{2}\)

\(\frac{A + B}{2}=\frac{3\alpha+6\alpha}{2}=\frac{9\alpha}{2}\) and \(\frac{A - B}{2}=\frac{3\alpha - 6\alpha}{2}=-\frac{3\alpha}{2}\).
Since \(\cos(-x)=\cos x\), we can use \(\cos\frac{3\alpha}{2}\) instead of \(\cos(-\frac{3\alpha}{2})\).

Step4: Substitute into the formula

\(\sin3\alpha+\sin6\alpha = 2\sin\frac{9\alpha}{2}\cos\frac{3\alpha}{2}\)

Answer:

\(2\sin\frac{9\alpha}{2}\cos\frac{3\alpha}{2}\)