Question

$\cos {40}^{\circ }+\cos {80}^{\circ }+\cos {160}^{\circ }+\cos {240}^{\circ }$

$=$

• A. $\frac{1}{2}$
  
• B. $-\frac{1}{2}$
  
• C. $1$
• D. $0$

Answer 1

The correct option is B $-\frac{1}{2}$


$\cos {40}^{\circ }+\cos {80}^{\circ }+\cos {160}^{\circ }+\cos {240}^{\circ }$

$[\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]$

$=2\cos \left(\frac{{40}^{\circ }+{80}^{\circ }}{2}\right)\cos \left(\frac{{40}^{\circ }-{80}^{\circ }}{2}\right)+\cos {160}^{\circ }+\cos ({180}^{\circ }+{60}^{\circ })$

$=2\cos \left({60}^{\circ }\right)\cos (-{20}^{\circ })+\cos {160}^{\circ }-\cos \left({60}^{\circ }\right)$

$=2\left(\frac{1}{2}\right)\cos (-{20}^{\circ })+\cos {160}^{\circ }-\frac{1}{2}$

$=\cos {20}^{\circ }+\cos {160}^{\circ }-\frac{1}{2}$

$=\cos ({180}^{\circ }-160{)}^{\circ }+\cos {160}^{\circ }-\frac{1}{2}$

$[\because \cos (180-\theta )=-\cos \theta ]$

$=-\cos {160}^{\circ }+\cos {160}^{\circ }-\frac{1}{2}$

$=-\frac{1}{2}$

Hence, Option (D) is correct.