Question

Prove that :

\[ \sin\frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \frac12 \]

Solution

Step 1: Reduce angles using periodicity:

\[ \sin\frac{13\pi}{3} = \sin\left(4\pi + \frac{\pi}{3}\right) = \sin\frac{\pi}{3} \]

\[ \sin\frac{2\pi}{3} \text{ is already in the first cycle.} \]

\[ \cos\frac{4\pi}{3} = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos\frac{\pi}{3} = -\frac12 \]

\[ \sin\frac{13\pi}{6} = \sin\left(2\pi + \frac{\pi}{6}\right) = \sin\frac{\pi}{6} = \frac12 \]

Step 2: Substitute the values:

\[ \sin\frac{13\pi}{3}\sin\frac{2\pi}{3} + \cos\frac{4\pi}{3}\sin\frac{13\pi}{6} = \sin\frac{\pi}{3}\sin\frac{2\pi}{3} + (-\frac12)(\frac12) \]

Step 3: Simplify each term:

\[ \sin\frac{\pi}{3}\sin\frac{2\pi}{3} = \frac{\sqrt3}{2} \cdot \frac{\sqrt3}{2} = \frac34 \]

\[ (-\frac12)(\frac12) = -\frac14 \]

Step 4: Add the two terms:

\[ \frac34 – \frac14 = \frac12 \]

Hence Proved.