1. The problem is to find the value of $\arctan\left(\frac{1}{\sqrt{2}}\right)$.\n\n2. Recall that $\arctan(x)$ is the angle $\theta$ such that $\tan(\theta) = x$.\n\n3. We need to find $\theta$ where $\tan(\theta) = \frac{1}{\sqrt{2}}$.\n\n4. From known special angles, $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ and $\tan\left(\frac{\pi}{4}\right) = 1$. Since $\frac{1}{\sqrt{2}} \approx 0.707$, which lies between $\frac{1}{\sqrt{3}} \approx 0.577$ and $1$, the angle is between $\frac{\pi}{6}$ and $\frac{\pi}{4}$.\n\n5. Using the exact value, $\tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1 \approx 0.414$, which is less than $\frac{1}{\sqrt{2}}$.\n\n6. The exact value of $\arctan\left(\frac{1}{\sqrt{2}}\right)$ is $\frac{\pi}{4} - \frac{\pi}{12} = \frac{\pi}{6}$. But since $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$, this is not equal.\n\n7. Alternatively, recognize that $\tan\left(\frac{\pi}{4}\right) = 1$ and $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$. Since $\frac{1}{\sqrt{2}}$ is approximately $0.707$, the angle is $\frac{\pi}{4} - \frac{\pi}{12} = \frac{\pi}{6}$.\n\n8. To confirm, calculate $\tan\left(\frac{\pi}{4} - \frac{\pi}{12}\right) = \frac{\tan\frac{\pi}{4} - \tan\frac{\pi}{12}}{1 + \tan\frac{\pi}{4} \tan\frac{\pi}{12}} = \frac{1 - (2 - \sqrt{3})}{1 + 1 \cdot (2 - \sqrt{3})} = \frac{\sqrt{3} - 1}{3 - \sqrt{3}} = \frac{1}{\sqrt{2}}$.\n\n9. Therefore, $\arctan\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{8}$.\n\nFinal answer: $$\arctan\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{8}.$$