Prove that: \[ \sin 38^\circ + \sin 22^\circ = \sin 82^\circ \]
Solution
Using the identity:
\[
\sin A + \sin B
=
2 \sin \frac{A+B}{2}
\cos \frac{A-B}{2}
\]
Taking
\[
A = 38^\circ,\qquad B = 22^\circ
\]
Then,
\[
\sin 38^\circ + \sin 22^\circ
=
2 \sin \frac{38^\circ+22^\circ}{2}
\cos \frac{38^\circ-22^\circ}{2}
\]
\[
=
2 \sin \frac{60^\circ}{2}
\cos \frac{16^\circ}{2}
\]
\[
=
2 \sin 30^\circ \cos 8^\circ
\]
\[
=
2 \times \frac{1}{2} \times \cos 8^\circ
\]
\[
=
\cos 8^\circ
\]
Using the identity:
\[
\cos \theta = \sin (90^\circ-\theta)
\]
\[
\cos 8^\circ
=
\sin (90^\circ-8^\circ)
=
\sin 82^\circ
\]
Hence,
\[
\boxed{
\sin 38^\circ + \sin 22^\circ
=
\sin 82^\circ
}
\]