Question

cos 40° + cos 80° + cos 160° + cos 240° =
(a) 0

(b) 1

(c) $\frac{1}{2}$

(d) $-\frac{1}{2}$

Answer 1

(d) $-\frac{1}{2}$

$$\begin{gathered} \cos 40°+\cos 80°+\cos 160°+\cos 240° \\ =2\cos \left(\frac{40°+80°}{2}\right)\cos \left(\frac{40°-80°}{2}\right)+\cos 160°-\cos \left(180°+60°\right)\left[\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right] \\ =2\cos 60°\cos \left(-20°\right)+\cos 160°-\frac{1}{2} \\ =2\times \frac{1}{2}\cos 20°+\cos 160°-\frac{1}{2} \\ =-\cos \left(180-20\right)°+\cos 160°-\frac{1}{2} \\ =-\frac{1}{2} \end{gathered}$$