Question

State true or false.

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

• A. True
• B. False

Answer 1

The correct option is B False

$\cos {20}^{\circ }\cos {40}^{\circ }\cos {60}^{\circ }\cos {80}^{\circ }$

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

$=\frac{1}{4}[\cos ({20}^{\circ }+{40}^{\circ })+\cos ({20}^{\circ }-{40}^{\circ })]\cos {80}^{\circ }$

$=\frac{1}{4}[\cos {60}^{\circ }+\cos {20}^{\circ }]\cos {80}^{\circ }$

$=\frac{1}{8}[2\cos {60}^{\circ }\cos {80}^{\circ }+2\cos {20}^{\circ }\cos {80}^{\circ }]$

$=\frac{1}{8}[2\times \frac{1}{2}\cos {80}^{\circ }+2\cos {20}^{\circ }\cos {80}^{\circ }]$

$=\frac{1}{8}[\cos {80}^{\circ }+\cos ({20}^{\circ }+{80}^{\circ })+\cos ({20}^{\circ }-{80}^{\circ })]$

$=\frac{1}{8}[\cos {80}^{\circ }+\cos {100}^{\circ }+\cos {60}^{\circ }]$

$=\frac{1}{8}[\cos ({180}^{\circ }-{100}^{\circ })+\cos {100}^{\circ }+\frac{1}{2}]$

$=\frac{1}{8}[-\cos {100}^{\circ }+\cos {100}^{\circ }+\frac{1}{2}]$

$=\frac{1}{8}\times \frac{1}{2}=\frac{1}{16}$

Hence the given solution in the question is false.