Question

find the exact value of the function.

$\tan \frac { \beta } { 2 }$, given $\tan \beta = \frac { \sqrt { 7 } } { 3 }$, with $180 ^ { \circ } < \beta < 270 ^ { \circ }$

$\tan \frac { \beta } { 2 } = \square$
(type an exact answer, using radicals as needed. rationalize all denominators.)

Explanation:

Step1: Determine the quadrant of \(\frac{\beta}{2}\)

Since \(180^{\circ}<\beta < 270^{\circ}\), then \(90^{\circ}<\frac{\beta}{2}<135^{\circ}\). So \(\frac{\beta}{2}\) is in the second quadrant, and \(\tan\frac{\beta}{2}<0\).

Step2: Use the half - angle formula for tangent

The half - angle formula for tangent is \(\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1 + \cos\alpha}\) or \(\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}\). Also, given \(\tan\beta=\frac{\sqrt{7}}{3}=\frac{\sin\beta}{\cos\beta}\), and \(\sin^{2}\beta+\cos^{2}\beta = 1\).
Since \(\tan\beta=\frac{\sqrt{7}}{3}\) and \(180^{\circ}<\beta<270^{\circ}\) (where \(\sin\beta<0\) and \(\cos\beta<0\)), we have \(\sin\beta=-\frac{\sqrt{7}}{\sqrt{(\sqrt{7})^{2}+3^{2}}}=-\frac{\sqrt{7}}{4}\), \(\cos\beta=-\frac{3}{4}\).
Using the formula \(\tan\frac{\beta}{2}=\frac{1-\cos\beta}{\sin\beta}\) (because of the sign in the second quadrant).
Substitute \(\sin\beta =-\frac{\sqrt{7}}{4}\) and \(\cos\beta=-\frac{3}{4}\) into the formula:

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Answer:

\(-\sqrt{7}\)