Question

The value of $\cos {50}^{\circ }+\cos {70}^{\circ }+\cos {170}^{\circ }$ is

• A. $-1$
• B. $-\frac{1}{2}$
• C. $0$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $0$

$\cos {50}^{\circ }+\cos {70}^{\circ }+\cos {170}^{\circ }$
$=2\cos \left(\frac{{50}^{\circ }+{70}^{\circ }}{2}\right)\cos \left(\frac{{50}^{\circ }-{70}^{\circ }}{2}\right)+\cos {170}^{\circ }$
$=2\cos {60}^{\circ }\cos (-{10}^{\circ })+\cos ({180}^{\circ }-{10}^{\circ })$
$=\cos {10}^{\circ }-\cos {10}^{\circ }=0$

Alternate Solution:
We know that,
$\cos \alpha +\cos (\alpha +\frac{2\pi }{3})+\cos (\alpha +\frac{4\pi }{3})=0$
Now, $\cos {50}^{\circ }+\cos {70}^{\circ }+\cos {170}^{\circ }$
$=\cos {50}^{\circ }+\cos {170}^{\circ }+\cos {70}^{\circ }$
$=\cos {50}^{\circ }+\cos (50+120{)}^{\circ }+\cos (360-70{)}^{\circ }$
$=\cos {50}^{\circ }+\cos (50+120{)}^{\circ }+\cos (50+240{)}^{\circ }$

Let $\alpha ={50}^{\circ }$
$\cos \alpha +\cos (\alpha +\frac{2\pi }{3})+\cos (\alpha +\frac{4\pi }{3})$
$=0$