# Solve an inequality $\sin t > \sin \left(t+\frac{\pi}{3}\right)$

I need to solve an inequality.

$$\sin t > \sin \left(t+\frac{\pi}{3}\right)$$

I simplified the inequality to:

$$\sin t > \sqrt3\ \cos t$$

I'm not sure what to do next. I know I can't square both sides, because I don't know whether they are positive or negative.

• Case distinction. a) $\cos t = 0$, easy; b) $\cos t > 0$, that yields $\tan t > \sqrt{3}$, and one has to weed out the intervals where $\cos t < 0$; c) $\cos t < 0$, that yields $\tan t < \sqrt{3}$, and one has again to weed out some intervals. – Daniel Fischer Oct 29 '13 at 15:23

I think there is a better way using $$\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2$$
$$\sin t-\sin\left(t+\frac\pi3\right)=-2\sin\frac\pi6\cos\left(t+\frac\pi6\right)=-\cos\left(t+\frac\pi6\right)$$
We need $\cos\left(t+\frac\pi6\right)<0$
$\displaystyle\implies 2n\pi+\frac\pi2<t+\frac\pi6<2n\pi+\pi+\frac\pi2$ where $n$ is any integer
as $\cos\theta<0$ in the second & the third quadrant