1. **Stating the problem:** We need to find the radian values of the given arcsin expressions and match them to the correct radian measures from the second group. 2. **Formula and rules:** Recall that $\arcsin(x)$ gives the angle $\theta$ in radians such that $\sin(\theta) = x$ and $\theta$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. 3. **Evaluate each arcsin expression:** - $\arcsin\left(\frac{\sqrt{2}}{2}\right)$: Since $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, this equals $\frac{\pi}{4}$. - $\arcsin\left(\frac{1}{2}\right)$: Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, this equals $\frac{\pi}{6}$. - $\arcsin(1)$: Since $\sin\left(\frac{\pi}{2}\right) = 1$, this equals $\frac{\pi}{2}$. - $\arcsin(0)$: Since $\sin(0) = 0$, this equals $0$. - $\arcsin\left(\frac{\sqrt{3}}{2}\right)$: Since $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$, this equals $\frac{\pi}{3}$. 4. **Summary of matches:** - $\arcsin\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$ - $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$ - $\arcsin(1) = \frac{\pi}{2}$ - $\arcsin(0) = 0$ - $\arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ These are the correct radian values corresponding to each arcsin expression. **Final answer:** $$\arcsin\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}, \quad \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}, \quad \arcsin(1) = \frac{\pi}{2}, \quad \arcsin(0) = 0, \quad \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}.$$