Question:
Show that sin 100° − sin 10° is positive.
Solution:
Let $f(\theta)=\sin 100^{\circ}-\sin 10^{\circ}$
Multiplying and dividing by $\sqrt{1^{2}+1^{2}}$, i.e. by $\sqrt{2}$, we get:
$\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin 100^{\circ}-\frac{1}{\sqrt{2}} \sin 10^{\circ}\right)$
$=\sqrt{2}\left(\cos 45^{\circ} \sin \left(90^{\circ}+10^{\circ}\right)-\sin 45^{\circ} \sin 10^{\circ}\right)$
$=\sqrt{2}\left(\cos 45^{\circ} \cos 10^{\circ}-\sin 45^{\circ} \sin 10^{\circ}\right)$
$=\sqrt{2} \cos \left(45^{\circ}+10^{\circ}\right)=\sqrt{2} \cos 55^{\circ}$, which is positive since $\cos$ is positive in the first quadrant.