Question

State whether the statement is true/false.

$\cos {36}^{\circ }.\cos {72}^{\circ }.\cos {108}^{\circ }.\cos {144}^{\circ }=\frac{1}{16}$

• A. True
• B. False

Answer 1

The correct option is A True

Given

LHS

$⟹\cos {36}^{\circ }\cos {72}^{\circ }\cos {108}^{\circ }\cos {144}^{\circ }$

Multiply and Divide by $2\sin {36}^{\circ }$

$⟹\frac{2\sin {36}^{\circ }}{2\sin {36}^{\circ }}\times \cos {36}^{\circ }\cos {72}^{\circ }\cos {108}^{\circ }\cos {144}^{\circ }$

$⟹\frac{1}{2\sin {36}^{\circ }}\times \sin {72}^{\circ }\cos {72}^{\circ }\cos {108}^{\circ }\cos {144}^{\circ }$

$⟹\frac{1}{4\sin {36}^{\circ }}\times \sin {144}^{\circ }\cos {108}^{\circ }\cos {144}^{\circ }$

$⟹\frac{1}{8\sin {36}^{\circ }}\times \sin 288\cos {108}^{\circ }$

$⟹\frac{-\sin 108\cos 108}{8\sin {36}^{\circ }}$

$⟹\frac{-\sin 216}{16\sin {36}^{\circ }}$

$⟹\frac{1}{16}\times \frac{\sin {36}^{\circ }}{\sin {36}^{\circ }}$

$⟹\frac{1}{16}$