Question

Calculate \[ \sin\frac{8\pi}{9}\, \sin\frac{7\pi}{9}\, \sin\frac{2\pi}{3}\, \sin\frac{5\pi}{9}. \]

• A. \( \frac34 \)
• B. \( \frac38 \)
• C. \( \frac{3}{16} \)
• D. \( \frac{3}{32} \)

Hint:

Memorize the important identity \[ \sin20^\circ\,\sin40^\circ\,\sin80^\circ = \frac{\sqrt3}{8}. \] Many trigonometric product questions reduce directly to this result.

Answer 1

The Correct Option is C

Concept: We use the identity \[ \sin(\pi-\theta)=\sin\theta \] to simplify the angles and then apply the standard product identity \[ \sin\theta\,\sin2\theta\,\sin4\theta = \frac14\sin4\theta. \]

Step 1: Convert all angles to acute angles. Using \[ \sin(\pi-\theta)=\sin\theta, \] we obtain \[ \sin\frac{8\pi}{9} = \sin\frac{\pi}{9}, \] \[ \sin\frac{7\pi}{9} = \sin\frac{2\pi}{9}, \] \[ \sin\frac{5\pi}{9} = \sin\frac{4\pi}{9}. \] Hence \[ P= \sin\frac{\pi}{9} \sin\frac{2\pi}{9} \sin\frac{4\pi}{9} \sin\frac{2\pi}{3}. \]

Step 2: Apply the product identity. Let \[ \theta=\frac{\pi}{9}. \] Then \[ \sin\frac{\pi}{9} \sin\frac{2\pi}{9} \sin\frac{4\pi}{9} = \frac14\sin\frac{4\pi}{9}. \] Using the standard identity \[ \sin\theta\sin2\theta\sin4\theta = \frac14\sin4\theta, \] and substituting \(\theta=\frac{\pi}{9}\), \[ = \frac14\sin\frac{4\pi}{9}. \] A more commonly used result is \[ \sin\frac{\pi}{9} \sin\frac{2\pi}{9} \sin\frac{4\pi}{9} = \frac{\sqrt3}{8}. \] Therefore, \[ P = \frac{\sqrt3}{8} \cdot \sin\frac{2\pi}{3}. \]

Step 3: Substitute the value of \(\sin\frac{2\pi}{3}\). Since \[ \sin\frac{2\pi}{3} = \frac{\sqrt3}{2}, \] we get \[ P = \frac{\sqrt3}{8} \cdot \frac{\sqrt3}{2}. \] \[ = \frac{3}{16}. \]

Step 4: State the final answer. Hence, \[ \boxed{\frac{3}{16}}. \] Therefore, the correct option is \(\boxed{(C)}\).