Question

use the unit circle to find the value of tan 2π and even or odd trigonometric functions to find the value of tan(-2π). select the correct choice below and fill in any answer boxes within your choice. a. tan 2π = (type an exact answer, using radicals as needed. simplify your answer. rationalize the denominator.) b. the expression is undefined.

Explanation:

Step1: Recall the unit circle and tangent definition

On the unit circle, a point corresponding to angle \(\theta\) is \((\cos\theta, \sin\theta)\). The tangent function is defined as \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). For \(\theta = 2\pi\), we find the coordinates of the point on the unit circle. The angle \(2\pi\) (or \(360^\circ\)) corresponds to the point \((1, 0)\) on the unit circle, so \(\cos(2\pi)=1\) and \(\sin(2\pi)=0\).

Step2: Calculate \(\tan(2\pi)\)

Using the definition of tangent, \(\tan(2\pi)=\frac{\sin(2\pi)}{\cos(2\pi)}\). Substituting \(\sin(2\pi) = 0\) and \(\cos(2\pi)=1\) into the formula, we get \(\tan(2\pi)=\frac{0}{1}=0\).

For \(\tan(- 2\pi)\), we use the even - odd property of the tangent function. The tangent function is an odd function? Wait, no, actually \(\tan(-\theta)=-\tan\theta\), but also, the period of the tangent function is \(\pi\), and \(y = \tan x\) has a period of \(\pi\), and \(2\pi\) is a multiple of \(2\pi\) (the period of sine and cosine, but tangent has period \(\pi\)). Also, \(-2\pi\) is coterminal with \(2\pi\) (since adding \(2\pi\) to \(- 2\pi\) gives \(0\), and adding \(2\pi\) again gives \(2\pi\)). So \(\tan(-2\pi)=\tan(2\pi)\) because the period of tangent is \(\pi\), and \(-2\pi = 2\pi-4\pi\), but more simply, \(\tan(-2\pi)=\frac{\sin(-2\pi)}{\cos(-2\pi)}\). Since \(\sin(-x)=-\sin x\) and \(\cos(-x)=\cos x\), \(\sin(-2\pi)=-\sin(2\pi) = 0\) and \(\cos(-2\pi)=\cos(2\pi)=1\), so \(\tan(-2\pi)=\frac{0}{1}=0\). But for the first part, \(\tan(2\pi)\):

Answer:

\(\tan(2\pi)=0\)