Prove that \(2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\frac{1}{2}\)
We use the product-to-sum identity to prove the given trigonometric expression.
Identity Used
\[
2\sin A\sin B=\cos(A-B)-\cos(A+B)
\]
Proof
\[
2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}
\]
Applying the identity:
\[
= \cos\left(\frac{5\pi}{12}-\frac{\pi}{12}\right)
-\cos\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)
\]
\[
= \cos\frac{4\pi}{12}-\cos\frac{6\pi}{12}
\]
\[
= \cos\frac{\pi}{3}-\cos\frac{\pi}{2}
\]
Using standard values:
\[
= \frac{1}{2}-0
\]
\[
= \frac{1}{2}
\]
Hence Proved
\[
2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\frac{1}{2}
\]