Prove that: \[ \cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0 \]
Solution
Using the identity:
\[
\cos A + \cos B
=
2 \cos \frac{A+B}{2}
\cos \frac{A-B}{2}
\]
Taking
\[
A = 20^\circ,\qquad B = 100^\circ
\]
Then,
\[
\cos 20^\circ + \cos 100^\circ
=
2 \cos \frac{20^\circ+100^\circ}{2}
\cos \frac{20^\circ-100^\circ}{2}
\]
\[
=
2 \cos 60^\circ \cos (-40^\circ)
\]
Since,
\[
\cos(-\theta)=\cos\theta
\]
\[
=
2 \cos 60^\circ \cos 40^\circ
\]
\[
=
2 \times \frac{1}{2} \times \cos 40^\circ
\]
\[
=
\cos 40^\circ
\]
Now,
\[
\cos 140^\circ
=
\cos(180^\circ-40^\circ)
=
-\cos 40^\circ
\]
Therefore,
\[
\cos 20^\circ + \cos 100^\circ + \cos 140^\circ
=
\cos 40^\circ – \cos 40^\circ
\]
\[
=0
\]
Hence,
\[
\boxed{
\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0
}
\]