Question

Prove that Cos 20+Cos 100+Cos 120=0

Answer 1

Correct Question:-

cos20° +cos100°+cos140°=0

SOLUTION :-

cos20°+cos100° It is in form of cosC+cosD

$cosC+cosD= 2cos\bigg(\dfrac{C+D}{2}\bigg) cos\bigg(\dfrac{C-D}{2} \bigg)$

$cos20^{\circ}+cos100^{\circ}= 2cos\bigg(\dfrac{20^{\circ}+100^{\circ}}{2} \bigg)cos\bigg(\dfrac{20^{\circ}-100^{\circ}}{2} \bigg)$

$cos20^{\circ}+cos100^{\circ}= 2cos\bigg(\dfrac{120^{\circ}}{2} \bigg)cos\bigg(\dfrac{-80^{\circ}}{2} \bigg)$

cos(-A) = cosA

so, cos(-40°) = cos40°

$cos20^{\circ}+cos100^{\circ}= 2cos60^{\circ} cos40^{\circ}$

$cos20^{\circ}+cos100^{\circ}= 2\frac{1}{2} cos40^{\circ}$

$cos20^{\circ}+cos100^{\circ}= cos40^{\circ}$

So, cos20° + cos100° = cos40°

$cos40^{\circ}+cos140^{\circ}$

Again it is in form of cosC+cosD by applying the formula we get

$cos40^{\circ}+cos140^{\circ}= 2cos\bigg(\dfrac{40^{\circ}+140^{\circ}}{2} \bigg)cos\bigg(\dfrac{40^{\circ}-140^{\circ}}{2} \bigg)$

$cos40^{\circ}+140^{\circ}= 2cos90^{\circ} \times cos50 ^{\circ}$

$cos40^{\circ}+cos140^{\circ}= 2(0) \times cos40^{\circ}$

$cos40^{\circ} + cos140^{\circ} = 0$

$cos20^{\circ} + cos100 ^{\circ}+ cos140^{\circ} = 0$

Hence proved !