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 \(\tan\frac{\alpha}{2}=\frac{1 - \cos\alpha}{\sin\alpha}\) or \(\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{1 + \cos\alpha}}\). Another form is \(\tan\frac{\beta}{2}=\frac{\sin\beta}{1+\cos\beta}\).
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 \(180^{\circ}<\beta<270^{\circ}\), we know that \(\sin\beta=-\frac{\sqrt{7}}{\sqrt{7 + 9}}=-\frac{\sqrt{7}}{4}\) and \(\cos\beta=-\frac{3}{4}\) (because in the third quadrant \(\sin\beta<0\) and \(\cos\beta<0\)).

Step3: Substitute into the half - angle formula

\(\tan\frac{\beta}{2}=\frac{\sin\beta}{1+\cos\beta}\)
Substitute \(\sin\beta =-\frac{\sqrt{7}}{4}\) and \(\cos\beta=-\frac{3}{4}\) into the formula:

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Answer:

\(-\sqrt{7}\)