Question

$\cos \left(1°\right)+\cos \left(2°\right)+\cos \left(3°\right)+·······+\cos \left(180°\right)$ is equal to

• A. $1$
• B. $0$
• C. $2$
• D. $-1$

Answer 1

The correct option is D

$-1$

Explanation for correct option:

Simplifying the series by Trigonometric identities:

Here, we use $\cos (\mathrm{\pi }-\mathrm{\theta })=-\mathrm{cos\theta }$

$\begin{array}{rcl}& & \cos \left(1°\right)+\cos \left(2°\right)+\cos \left(3°\right)+·······+\cos \left(180°\right)\\ & =& \cos \left(1°\right)+\cos \left(2°\right)+\cos \left(3°\right)+·····+\cos \left(90°\right)+·····+\cos \left(177°\right)+\cos \left(178°\right)+\cos \left(179°\right)+\cos \left(180°\right)\\ & =& \cos \left(1°\right)+\cos \left(2°\right)+\cos \left(3°\right)+·····+\cos \left(90°\right)+·····+\cos \left(177°\right)+\cos \left(178°\right)+\cos \left(179°\right)+\cos \left(180°\right)\\ & =& \cos \left(1°\right)+\cos \left(2°\right)+\cos \left(3°\right)+·····+\cos \left(90°\right)+·····+\cos \left(180°-3°\right)+\cos \left(180°-2°\right)+\cos \left(180°-1°\right)+\cos \left(180°\right)\\ & =& \cos \left(1°\right)+\cos \left(2°\right)+\cos \left(3°\right)+·····+\cos \left(90°\right)+·····-\cos \left(1°\right)-\cos \left(2°\right)-\cos \left(1°\right)+\cos \left(180°\right)\\ & =& \cos \left(90°\right)+\cos \left(180°\right)\\ & =& 0-1\\ & =& -1\end{array}$

Therefore, Option (D) is correct.