Question

$\cos 1^o+\cos 2^o+\cos 3^o+...+\cos {180}^o$ is equal to

• A. 1
• B. 0
• C. 2
• D. -1

Answer 1

The correct option is D -1

$\cos 1^o+\cos 2^o+....+\cos {180}^o$

Using the identity $\cos (180-\theta )=-\cos \theta$, we can write the above expression as

$\cos 1^o+\cos 2^o+...+\cos {90}^o-\cos {89}^o-\cos {88}^o-...-\cos 1^o+\cos {180}^o$

$=\cos {90}^o+\cos {180}^o$

$=0-1=-1$