Question

Prove that $cos{20}^{\circ }cos{40}^{\circ }cos{60}^{\circ }cos{80}^{\circ }=\frac{1}{16}$

Answer 1

$\cos 20°\cos 40°\cos 60°\cos 80°=\frac{1}{16}$

$\Rightarrow LHS=\cos 20°\cos 40°\cos 60°\cos 80°$

$=\frac{1}{2}\left(\cos 20°\cos 40°\cos 80°\right)$.............$cos{60}^{\circ }=\frac{1}{2}$

Let $20°=\theta ,$ then $40°=2\theta \&80°=4\theta$

$=\frac{1}{2}\cos \theta \cos 2\theta \cos 4\theta$

$=\frac{1}{2}\left(\frac{\sin 2^3\theta }{2^3\sin \theta }\right)$...........$cos\theta .cos2\theta .cos{2}^{n-1}\theta =\frac{sin{2}^n\theta }{2^{n}sin\theta }$

$=\frac{1}{24}\left(\frac{\sin 8\theta }{\sin \theta }\right)$

$=\frac{1}{16}\frac{\sin 160°}{\sin 20°}$

$=\frac{1}{16}\frac{\sin (180°-20°)}{\sin 20°}$

$=\frac{1}{16}\frac{\sin 20°}{\sin 20°}$

$=\frac{1}{16}$

$LHS=RHS$

Hence, it proved.