Question

Solve:
$\cos {40}^0\cos {100}^0\cos {160}^0$

Answer 1

$\cos {40}^{\circ }\cos {100}^{\circ }\cos {160}^{\circ }$

$=\cos {40}^{\circ }\cos ({180}^{\circ }-{80}^{\circ })\cos ({180}^{\circ }-{20}^{\circ })$

$=\cos {40}^{\circ }(-\cos {80}^{\circ })(-\cos {20}^{\circ })$

$=\cos {20}^{\circ }\cos {40}^{\circ }\cos {80}^{\circ }$

$=\frac{1}{2\sin {20}^{\circ }}\times 2\sin {20}^{\circ }\cos {20}^{\circ }\cos {40}^{\circ }\cos {80}^{\circ }$

$=\frac{1}{4\sin {20}^{\circ }}\times 2\sin {40}^{\circ }\cos {40}^{\circ }\cos {80}^{\circ }$

$=\frac{1}{8\sin {20}^{\circ }}\times 2\sin {80}^{\circ }\cos {80}^{\circ }$

$=\frac{1}{8\sin {20}^{\circ }}\times \sin {160}^{\circ }$

$=\frac{1}{8\sin {20}^{\circ }}\times \sin ({180}^{\circ }-{20}^{\circ })$

$=\frac{1}{8\sin {20}^{\circ }}\times \sin {20}^{\circ }$

$=\frac{1}{8}$