Question

Prove that:

(iii) $\cos {80}^0+\cos {40}^0-\cos {20}^0=0$

Answer 1

Now let's prove,

$\cos {80}^0+\cos {40}^0-\cos {20}^0=0$

$L.H.S.=\cos {80}^0+\cos {40}^0-\cos {20}^0$

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

$=2\cos \left(\frac{{80}^0+{40}^0}{2}\right)\cos \left(\frac{{80}^0-{40}^0}{2}\right)-\cos {20}^0$

$=2\cos {60}^0\cos {20}^0-\cos {20}^0$

$=2\times \frac{1}{2}\cos {20}^0-\cos {20}^0[\because \cos {60}^0=\frac{1}{2}]$

$=\cos {20}^0-\cos {20}^0$

$=0$

$=R.H.S$

Hence, proved.