Question

prove that:

cos20°+cos100°+cos120°=0

help me with this question and needed proper solution if any one misbehave here i will report the user and if you give me proper answer you will get 100 points​

Answer 1

Step-by-step explanation:

Use formula cos(a+b) = cosAsinB - sinAcosB (pls confirm)

Answer 2

$\huge\fbox{Answer ☘}$

Question -

cos 20° + cos 100° + cos 140° = 0

Solution~

consider ( cos 20° and cos 100° )

these are in the form of cos C + cos D

cosC + cosD =

$2 \cos( \frac{c + d}{2} ) \cos( \frac{c - d}{2} )$

cos 20° + cos 100° =

$2 \cos( \frac{20 + 100}{2} ) \cos( \frac{20 - 100}{2} ) \\ \\ 2 \cos( \frac{120}{2} ) \cos( \frac{ - 80}{2} )$

Now ,

cos ( - A ) = cos A

so cos(-40°) = cos40°

cos20° + cos100° = 2 cos60°cos40°

$\cos(20) + \cos(100) = 2 \frac{1}{2} \cos(40) \\ \\ \cos(20) + \cos(100) = \cos(40)$

so ,

cos20° + cos100° = cos40°

$\cos(40) + \cos(140)$

Again it is in the form of cosC + cosD

by applying this formula , we get

$\cos(40) + \cos(140) = \\ 2 \cos( \frac{40 + 140}{2} ) \cos( \frac{40 - 140}{2} ) \\ \\ \cos(40) + \cos(140) = 2 \cos(90) \times \cos(50)$

$\cos(40) + \cos(140) = 2(0) \times \cos(40) \\ \\ \cos(40) + \cos(140) = 0 \\ \\ \cos(20) + \cos(100) + \cos(140) = 0$

hence , proved!

hope helpful~