Prove that: \[ \cos 80^\circ + \cos 40^\circ – \cos 20^\circ = 0 \]
Solution
Using the identity:
\[
\cos A + \cos B
=
2 \cos \frac{A+B}{2}
\cos \frac{A-B}{2}
\]
Taking
\[
A = 80^\circ,\qquad B = 40^\circ
\]
Then,
\[
\cos 80^\circ + \cos 40^\circ
=
2 \cos \frac{80^\circ+40^\circ}{2}
\cos \frac{80^\circ-40^\circ}{2}
\]
\[
=
2 \cos 60^\circ \cos 20^\circ
\]
\[
=
2 \times \frac{1}{2} \times \cos 20^\circ
\]
\[
=
\cos 20^\circ
\]
Therefore,
\[
\cos 80^\circ + \cos 40^\circ – \cos 20^\circ
=
\cos 20^\circ – \cos 20^\circ
\]
\[
=0
\]
Hence,
\[
\boxed{
\cos 80^\circ + \cos 40^\circ – \cos 20^\circ = 0
}
\]