Question

Prove that:

(i) $\cos {55}^{\circ }+\cos {65}^{\circ }+\cos {175}^{\circ }=0$

Answer 1

Now let's prove, $\cos {55}^{\circ }+\cos {65}^{\circ }+\cos {175}^{\circ }=0$

$L.H.S.=\cos {55}^{\circ }+\cos {65}^{\circ }+\cos {175}^{\circ }$

$[\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)]$

$=2\cos \left(\frac{{55}^{\circ }+{65}^{\circ }}{2}\right)\cos \left(\frac{{55}^{\circ }-{65}^{\circ }}{2}\right)+\cos {175}^{\circ }$

$=2\cos {60}^{\circ }\cos (-5{)}^{\circ }+\cos {175}^{\circ }$

$=2\times \frac{1}{2}\cos 5^{\circ }+\cos {175}^{\circ }$

$=\cos 5^{\circ }+\cos {175}^{\circ }$

$=2\cos \left(\frac{5^{\circ }+{175}^{\circ }}{2}\right)\times \cos \left(\frac{5^{\circ }-{175}^{\circ }}{2}\right)$

$=2\cos {90}^{\circ }\times \cos {85}^{\circ }[\because \cos {90}^{\circ }=0]$

$=0$

$=R.H.S.$

Hence, proved.