Question

$\cos {20}^{\circ }\cos {100}^{\circ }+\cos {100}^{\circ }\cos {140}^{\circ }-\cos {140}^{\circ }.\cos {200}^{\circ }$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $-\frac{1}{4}$
• D. $-\frac{3}{4}$

Answer 1

The correct option is D $-\frac{3}{4}$

$\cos {20}^{\circ }\cos {100}^{\circ }+\cos {100}^{\circ }\cos {140}^{\circ }-\cos {140}^{\circ }\cos {200}^{\circ }$
$=\frac{1}{2}(\cos {120}^{\circ }+\cos {80}^{\circ }+\cos {240}^{\circ }+\cos {40}^{\circ }-\cos {340}^{\circ }-\cos {60}^{\circ })$
$(\because \cos A\cos B=\frac{1}{2}[\cos (A+B)+\cos (A-B)\left]\right)$
$=\frac{1}{2}(-\frac{1}{2}+\cos {80}^{\circ }-\frac{1}{2}+\cos {40}^{\circ }-\cos {340}^{\circ }-\frac{1}{2})(\because \cos {60}^{\circ }=\frac{1}{2})$
$=\frac{1}{2}(-\frac{3}{2}+\cos {80}^{\circ }+\cos {40}^{\circ }-\cos {340}^{\circ })$
$=\frac{1}{2}(-\frac{3}{2}+2\cos {60}^{\circ }\cos {20}^{\circ }-\cos {340}^{\circ })$
$=\frac{1}{2}(-\frac{3}{2}+\cos {20}^{\circ }-\cos {340}^{\circ })$
$=\frac{1}{2}(-\frac{3}{2}+\cos {20}^{\circ }-\cos {({360}^{\circ }-20}^{\circ }))$
$=\frac{1}{2}(-\frac{3}{2}+\cos {20}^{\circ }-\cos {(-20}^{\circ }))$
$=\frac{1}{2}(-\frac{3}{2}+0)$
$=-\frac{3}{4}$

Hence, $\cos {20}^{\circ }\cos {100}^{\circ }+\cos {100}^{\circ }\cos {140}^{\circ }-\cos {140}^{\circ }\cos {200}^{\circ }=-\frac{3}{4}$.