Question

The value of cos 52° + cos 68° + cos 172° is
(a) 0
(b) 1
(c) 2
(d) 3/2

Answer 1

(a) 0

$$\begin{gathered} \cos 52°+\cos 68°+\cos 172° \\ =2\cos \left(\frac{52°+68°}{2}\right)\cos \left(\frac{52°-68°}{2}\right)+\cos 172°\left[\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right] \\ =2\cos 60°\cos \left(-8°\right)+\cos 172° \\ =2\times \frac{1}{2}\cos 8°+\cos 172° \\ =\cos 8°+\cos 172° \\ =2\cos \left(\frac{8°+172°}{2}\right)\cos \left(\frac{8°-172°}{2}\right) \\ =2\cos 90°\cos 82° \\ =0 \end{gathered}$$