Question

1. arctan\frac{\sqrt{3}}{3}

Explanation:

Step1: Recall the definition of arctangent function

The function \(y = \arctan(x)\) is the inverse function of \(y=\tan(x)\) with the domain \(x\in(-\frac{\pi}{2},\frac{\pi}{2})\). We need to find an angle \(\theta\in(-\frac{\pi}{2},\frac{\pi}{2})\) such that \(\tan\theta=\frac{\sqrt{3}}{3}\).

Step2: Use the values of tangent function for special angles

We know that \(\tan\frac{\pi}{6}=\frac{\sin\frac{\pi}{6}}{\cos\frac{\pi}{6}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\), and \(\frac{\pi}{6}\in(-\frac{\pi}{2},\frac{\pi}{2})\)

Answer:

\(\frac{\pi}{6}\)