Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $179$

Answer 1

The correct option is A $-1$

$\Rightarrow \cos 1^o+\cos 2^o+\cos 3^o......+\cos {180}^o$

$=\cos 1^o+\cos 2^o+.....+\cos {178}^o+\cos {179}^o+cos{180}^o$

$=\cos 1^o+\cos 2^o+.....+\cos ({178}^o-2^o)+\cos ({179}^o-1^o)+cos({180}^o-0)$

$=\cos 1^o+\cos 2^o.....-\cos 2^o-\cos 1^o-\cos 0$

$\because \left(\cos \right(180-\theta )=-\cos {\theta }^o)$

$=-\cos 0$

$=-1$