Question

If $\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3=3$, then $\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3=$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $\left(0\right)$

Given, $\sin {\theta }_1+\sin {\theta }_2+\sin {\theta }_3=3$
This value is possible only if $\sin {\theta }_1=\sin {\theta }_2=\sin {\theta }_3=1$
Since $\sin \frac{\pi }{2}=1$
So, $\sin {\theta }_1=\sin \frac{\pi }{2}$
Therefore, ${\theta }_1={\theta }_2={\theta }_3=\frac{\pi }{2}$
Now, substitute the values of ${\theta }_1,{\theta }_2,{\theta }_3$ in $\cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3$, we get
$\cos \frac{\pi }{2}+\cos \frac{\pi }{2}+\cos \frac{\pi }{2}=0+0+0$
$\therefore \cos {\theta }_1+\cos {\theta }_2+\cos {\theta }_3=0$