Question

use the product - to - sum identities to rewrite the expression.
2\cos\left(\frac{\pi}{5}\
ight)\sin\left(\frac{\pi}{9}\
ight)
which choice below is equivalent to 2\cos\left(\frac{\pi}{5}\
ight)\sin\left(\frac{\pi}{9}\
ight)?
\\( \bigcirc \\) a. \\( \sin \frac{14 \pi}{45}-\sin \frac{4 \pi}{45} \\)
\\( \bigcirc \\) b. \\( \sin \frac{14 \pi}{45}+\sin \frac{4 \pi}{45} \\)
\\( \bigcirc \\) c. \\( 2\left(\sin \frac{14 \pi}{45}-\sin \frac{4 \pi}{45}\
ight) \\)
\\( \bigcirc \\) d. \\( \cos \frac{14 \pi}{45}-\cos \frac{4 \pi}{45} \\)

Explanation:

Step1: Recall the product - to - sum identity

The identity \(2\cos A\sin B=\sin(A + B)-\sin(A - B)\).

Step2: Substitute \(A=\frac{\pi}{5}\) and \(B = \frac{\pi}{9}\)

First, calculate \(A + B=\frac{\pi}{5}+\frac{\pi}{9}=\frac{9\pi+5\pi}{45}=\frac{14\pi}{45}\).
Second, calculate \(A - B=\frac{\pi}{5}-\frac{\pi}{9}=\frac{9\pi - 5\pi}{45}=\frac{4\pi}{45}\).
Then \(2\cos\frac{\pi}{5}\sin\frac{\pi}{9}=\sin(\frac{\pi}{5}+\frac{\pi}{9})-\sin(\frac{\pi}{5}-\frac{\pi}{9})=\sin\frac{14\pi}{45}-\sin\frac{4\pi}{45}\).

Answer:

A. \(\sin\frac{14\pi}{45}-\sin\frac{4\pi}{45}\)