Prove that: \[ \sin 50^\circ – \sin 70^\circ + \sin 10^\circ = 0 \]
Solution
Using the identity:
\[
\sin A – \sin B
=
2 \cos \frac{A+B}{2}
\sin \frac{A-B}{2}
\]
Taking
\[
A = 50^\circ,\qquad B = 70^\circ
\]
Then,
\[
\sin 50^\circ – \sin 70^\circ
=
2 \cos \frac{50^\circ+70^\circ}{2}
\sin \frac{50^\circ-70^\circ}{2}
\]
\[
=
2 \cos 60^\circ \sin (-10^\circ)
\]
Since,
\[
\sin(-\theta)=-\sin\theta
\]
\[
=
-2 \cos 60^\circ \sin 10^\circ
\]
\[
=
-2 \times \frac{1}{2} \times \sin 10^\circ
\]
\[
=
-\sin 10^\circ
\]
Therefore,
\[
\sin 50^\circ – \sin 70^\circ + \sin 10^\circ
=
-\sin 10^\circ + \sin 10^\circ
\]
\[
=0
\]
Hence,
\[
\boxed{
\sin 50^\circ – \sin 70^\circ + \sin 10^\circ = 0
}
\]