Question

Prove that:

(iv) $\cos {20}^0+\cos {100}^0+\cos {140}^0=0$

Answer 1

Now let's prove,

$\cos {20}^0+\cos {100}^0+\cos {140}^0=0$

$L.H.S.=(\cos {20}^0+\cos {100}^0)+\cos {140}^0$

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

$=2\cos \left(\frac{{20}^0+{100}^0}{2}\right)\cos \left(\frac{{20}^0-{100}^0}{2}\right)+\cos {140}^0$

$=2\cos {60}^0\cos (-40{)}^0+\cos {140}^0$

$=2\times \frac{1}{2}\cos {40}^0+\cos {140}^0[\because \cos {60}^0=\frac{1}{2}\&\cos (-\theta )=\cos (\theta \left)\right]$

$=\cos {40}^0+\cos ({180}^0-{40}^0)$

$=\cos {40}^0-\cos {40}^0[\because \cos ({180}^0-\theta )=-\cos \theta ]$

$=0$

$=R.H.S.$

Hence, proved.