Question

prove that : Cos 20° + Cos 100° + Cos 140º = 0​

Answer 1

Step-by-step explanation:

cos20∘+cos100∘+cos140∘

=(cos20∘+cos140∘)+cos(90∘+10∘)

=2cos80∘cos60∘−sin10∘=cos80∘−sin10∘

=sin10∘−sin10∘=0

Answer 2

Answer:

$\huge\tt\underline{question}$

prove that : Cos 20° + Cos 100° + Cos 140º = 0

$\huge\tt\underline{answer}$

Consider LHS,

$\cos20° + \cos100° + \cos140°$

$= 2 \cos( \frac{20° + 100°}{2} ) \cos( \frac{20° - 100°}{2} ) + \cos(140°)$

$= 2 \cos60° \cos( - 140°) + \cos140°$

$= 2 \times \frac{1}{2} \cos40° + \cos140°$

$= \cos40° + \cos140°$

$= 2 \cos( \frac{40° + 140°}{2} ) \cos( \frac{40° - 140°}{2} )$

$= 2 \cos90° \cos50°$

$= 0$

$\bf\underline\pink{hence\:LHS\:=\:RHS}$