Compute $\arcsin \left( -\frac{1}{2} \right).$  Express your answer in radians.
I need to find the angle whose sine is $-\frac{1}{2}.$ I recall that the sine function is periodic with period $2\pi,$ so there are infinitely many such angles, but I only need one of them. I also know that the sine function is odd, meaning that $\sin(-x) = -\sin(x)$ for any angle $x.$ This means that if $x$ is an angle whose sine is $\frac{1}{2},$ then $-x$ is an angle whose sine is $-\frac{1}{2}.$ So I can focus on finding an angle whose sine is $\frac{1}{2}$ first, and then take its negative. I remember that the special right triangles with angles $30^\circ, 60^\circ,$ and $90^\circ$ have nice ratios for the sine, cosine, and tangent functions. In particular, the triangle with hypotenuse $2$ and legs $1$ and $\sqrt{3}$ has a $30^\circ$ angle opposite the side of length $1.$ This means that $\sin(30^\circ) = \frac{1}{2}.$ Therefore, $30^\circ$ is an angle whose sine is $\frac{1}{2},$ and $-30^\circ$ is an angle whose sine is $-\frac{1}{2}.$ However, the problem asks for the answer in radians, not degrees. I recall that $\pi$ radians is equal to $180^\circ,$ so I can convert degrees to radians by multiplying by $\frac{\pi}{180}.$ Thus, $-30^\circ$ in radians is $-\frac{30 \pi}{180} = -\frac{\pi}{6}.$ 