Trigonometrical Ratios and Identities 1 Question 26
26. Show that $16 \cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=1$
$(1983,2 M)$
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Solution:
- $16 \cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}$
$$ =16\left(\cos A \cdot \cos 2 A \cos 2^{2} A \cdot \cos 2^{3} A\right) $$
where, $A=\frac{2 \pi}{15}$
$$ =16 \frac{\sin 2^{4} A}{2^{4} \sin A}=\frac{\sin 2^{4} \frac{2 \pi}{15}}{\sin \frac{2 \pi}{15}} $$
$=\frac{\sin \frac{32 \pi}{15}}{\sin \frac{2 \pi}{15}}=\frac{\sin 2 \pi+\frac{2 \pi}{15}}{\sin \frac{2 \pi}{15}}$
$=\frac{\sin \frac{2 \pi}{15}}{\sin \frac{2 \pi}{15}}=1$
