Q117 / 160 · TS EAMCET 2003 · Mathematics
The value of $\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$ is :
- Correct answerA$\frac{1}{16}$
- B$\frac{1}{8}$
- C$\frac{1}{12}$
- D$\frac{1}{4}$
Solution
$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15}$
$=\cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ} \cos 168^{\circ}$
$=-\cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}$
$=-\frac{16 \sin 12^{\circ}}{16 \sin 12^{\circ}} \cos 12^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}$
$=\frac{-8 \sin 24^{\circ} \cos 24^{\circ} \cos 48^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}}$
$=\frac{-4 \sin 48^{\circ} \cos 48^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}}$
$=-\frac{2 \sin 96^{\circ} \cos 96^{\circ}}{16 \sin 12^{\circ}}$
$=-\frac{\sin 192^{\circ}}{16 \sin 12^{\circ}}=\frac{\sin 12^{\circ}}{16 \sin 12^{\circ}}=\frac{1}{16}$
Hence, the correct answer is option: 'A'.