### Instruction: Compute $\sin 150^\circ$.### Response: To solve this problem, I will use the fact that $\sin (A+B) = \sin A \cos B + \cos A \sin B$.
I will write $150^\circ$ as $90^\circ + 60^\circ$ and apply the formula.
I get $\sin 150^\circ = \sin (90^\circ + 60^\circ) = \sin 90^\circ \cos 60^\circ + \cos 90^\circ \sin 60^\circ$.
Now, I recall the values of the sine and cosine functions for some common angles.
I know that $\sin 90^\circ = 1$, $\cos 90^\circ = 0$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, and $\cos 60^\circ = \frac{1}{2}$.
Plugging these values into the expression, I simplify and obtain $\sin 150^\circ = 1 \cdot \frac{1}{2} + 0 \cdot \frac{\sqrt{3}}{2} = \frac{1}{2}$.