Calculate $\tan \frac{\pi}{6}$.
To determine the value of \(\tan \frac{\pi}{6}\), we first analyze the trigonometric definition of the tangent function. The tangent of an angle \(\theta\) is defined as the ratio of the sine of the angle to its cosine:
\[\tan \theta = \frac{\sin \theta}{\cos \theta}\]
In this problem, the angle is given in radians as \(\theta = \frac{\pi}{6}\), which is equivalent to \(30^\circ\) in degrees.
We can find the values of \(\sin \frac{\pi}{6}\) and \(\cos \frac{\pi}{6}\) by considering a right-angled triangle with angles \(30^\circ\), \(60^\circ\), and \(90^\circ\). In such a triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\). Specifically, for the angle \(30^\circ\):
Now, we substitute these values into the formula for the tangent:
\[\tan \frac{\pi}{6} = \frac{\sin \frac{\pi}{6}}{\cos \frac{\pi}{6}} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\]
By simplifying this complex fraction, we get:
\[\tan \frac{\pi}{6} = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}\]
To express the result in a standard form, we rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{3}\):
\[\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3}\]
Comparing this result to the given options, we find that the correct value is \(\frac{\sqrt{3}}{3}\).
Trigonometric functions like tangent are essential in surveying and architecture. For example, if you know the angle of elevation to the top of a building is \(\frac{\pi}{6}\) (or \(30^\circ\)) and the distance to the base of the building is \(x\), you can use the tangent function to determine the building's height \(h\): \(\tan(30^\circ) = \frac{h}{x}\). Knowing that \(\tan(30^\circ) = \frac{\sqrt{3}}{3}\), you can easily calculate that \(h = x \cdot \frac{\sqrt{3}}{3}\). This simple calculation allows builders and surveyors to measure heights of structures without needing to climb them.