Question

Prove that: $\cos {10}^0\cos {30}^0\cos {50}^0\cos {70}^0=\frac{3}{16}$

Answer 1

We know that-

$\cos A\cos (\frac{\pi }{3}+A)\cos (\frac{\pi }{3}-A)=\frac{1}{4}\cos 3A$

here $A=10$

$\Rightarrow \cos 10.\cos (60+10).\cos (60-10)=\frac{1}{4}\cos 3\left(10\right)$

$\Rightarrow \cos 10.\cos \left(70\right).\cos \left(30\right)=\frac{1}{4}\cos {30}^0$

Multiplying ${\cos 30}^0$ on both sides

$\Rightarrow {\cos 10}^0{\cos 30}^0.{\cos 50}^0\cos {70}^0=\frac{1}{4}{\cos }^2{30}^0$

$(\because {\cos 30}^0=\frac{\sqrt{3}}{2})$ $=\frac{1}{4}{\left(\frac{\sqrt{3}}{2}\right)}^2$

$=\frac{1}{4}\times \frac{3}{4}=\frac{3}{16}$

Hence proved