Prove that: \[ \sin 40^\circ + \sin 20^\circ = \cos 10^\circ \]
Solution
Using the identity:
\[
\sin A + \sin B
=
2 \sin \frac{A+B}{2}
\cos \frac{A-B}{2}
\]
Taking
\[
A = 40^\circ,\qquad B = 20^\circ
\]
Then,
\[
\sin 40^\circ + \sin 20^\circ
=
2 \sin \frac{40^\circ+20^\circ}{2}
\cos \frac{40^\circ-20^\circ}{2}
\]
\[
=
2 \sin \frac{60^\circ}{2}
\cos \frac{20^\circ}{2}
\]
\[
=
2 \sin 30^\circ \cos 10^\circ
\]
\[
=
2 \times \frac{1}{2} \times \cos 10^\circ
\]
\[
=
\cos 10^\circ
\]
Hence,
\[
\boxed{
\sin 40^\circ + \sin 20^\circ
=
\cos 10^\circ
}
\]