You should know that $\displaystyle \displaystyle \sin{\left(180^{\circ} - \theta^{\circ}\right)} \equiv \sin{\theta^{\circ}}$ for $\displaystyle \displaystyle 0^{\circ} < \theta^{\circ} < 90^{\circ}$.
Although the other two methods are easier you can use the fact that $\displaystyle \sin(A+B) = \sin A \cos B + \cos A \sin B \text{ where } A = 90^{\circ} \text{ and } B = 30^{\circ}$
30 and 90 are, of course, found on a 30-60-90 triangle