Prove that: \[ \sin 23^\circ + \sin 37^\circ = \cos 7^\circ \]
Solution
Using the identity:
\[
\sin A + \sin B
=
2 \sin \frac{A+B}{2}
\cos \frac{A-B}{2}
\]
Taking
\[
A = 23^\circ,\qquad B = 37^\circ
\]
Then,
\[
\sin 23^\circ + \sin 37^\circ
=
2 \sin \frac{23^\circ+37^\circ}{2}
\cos \frac{23^\circ-37^\circ}{2}
\]
\[
=
2 \sin \frac{60^\circ}{2}
\cos \frac{-14^\circ}{2}
\]
\[
=
2 \sin 30^\circ \cos (-7^\circ)
\]
Since,
\[
\cos(-\theta)=\cos\theta
\]
\[
=
2 \sin 30^\circ \cos 7^\circ
\]
\[
=
2 \times \frac{1}{2} \times \cos 7^\circ
\]
\[
=
\cos 7^\circ
\]
Hence,
\[
\boxed{
\sin 23^\circ + \sin 37^\circ
=
\cos 7^\circ
}
\]