Prove that: \[ \cos 100^\circ + \cos 20^\circ = \cos 40^\circ \]
Solution
Using the identity:
\[
\cos A + \cos B
=
2 \cos \frac{A+B}{2}
\cos \frac{A-B}{2}
\]
Taking
\[
A = 100^\circ,\qquad B = 20^\circ
\]
Then,
\[
\cos 100^\circ + \cos 20^\circ
=
2 \cos \frac{100^\circ+20^\circ}{2}
\cos \frac{100^\circ-20^\circ}{2}
\]
\[
=
2 \cos \frac{120^\circ}{2}
\cos \frac{80^\circ}{2}
\]
\[
=
2 \cos 60^\circ \cos 40^\circ
\]
\[
=
2 \times \frac{1}{2} \times \cos 40^\circ
\]
\[
=
\cos 40^\circ
\]
Hence,
\[
\boxed{
\cos 100^\circ + \cos 20^\circ
=
\cos 40^\circ
}
\]