Answer
$\sin 130^\circ-\sin 110^\circ=-\sin 10^\circ$
Work Step by Step
Start with the left side:
$\sin 130^\circ-\sin 110^\circ$
Use the Sum-to-Product Formulas on page 560:
$=2\cos\frac{130^\circ+110^\circ}{2}\sin\frac{130^\circ-110^\circ}{2}$
Simplify:
$=2\cos\frac{240^\circ}{2}\sin\frac{20^\circ}{2}$
$=2\cos 120^\circ\sin 10^\circ$
$=2*(-\frac{1}{2})\sin 10^\circ$
$=-\sin 10^\circ$
Since this equals the right side, the identity has been proven.
