Question

if \\(cos x = \frac{1}{5}\\), then what is the positive value of \\(\tan \frac{1}{2}x\\), in simplest radical form with a rational denominator?

Explanation:

Step 1: Recall the half - angle formula for tangent

The half - angle formula for tangent is \(\tan\frac{\theta}{2}=\frac{1 - \cos\theta}{\sin\theta}\) or \(\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1 + \cos\theta}}\) (we use the positive square root since we want the positive value of \(\tan\frac{1}{2}x\)). We know that \(\cos x=\frac{1}{5}\), and we can also use the identity \(\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}\) or derive it from the basic trigonometric identities. Another way is to use the formula \(\tan\frac{\alpha}{2}=\frac{1 - \cos\alpha}{\sin\alpha}\), and we can find \(\sin\alpha\) using the Pythagorean identity \(\sin^{2}\alpha+\cos^{2}\alpha = 1\).

First, find \(\sin x\). Since \(\sin^{2}x=1-\cos^{2}x\), substitute \(\cos x = \frac{1}{5}\) into it:
\(\sin^{2}x=1 - (\frac{1}{5})^{2}=1-\frac{1}{25}=\frac{24}{25}\), so \(\sin x=\pm\frac{2\sqrt{6}}{5}\). But since we are looking for the positive value of \(\tan\frac{x}{2}\), and we know that \(\tan\frac{x}{2}=\frac{1 - \cos x}{\sin x}\), we need to consider the sign of \(\sin x\). However, we can also use the formula \(\tan\frac{x}{2}=\sqrt{\frac{1 - \cos x}{1+\cos x}}\) (we take the positive root because we want the positive value of \(\tan\frac{x}{2}\)).

Step 2: Substitute \(\cos x=\frac{1}{5}\) into the half - angle formula for tangent

Using the formula \(\tan\frac{x}{2}=\sqrt{\frac{1-\cos x}{1 + \cos x}}\), substitute \(\cos x=\frac{1}{5}\):
\(1-\cos x=1-\frac{1}{5}=\frac{4}{5}\)
\(1 + \cos x=1+\frac{1}{5}=\frac{6}{5}\)
Then \(\frac{1-\cos x}{1 + \cos x}=\frac{\frac{4}{5}}{\frac{6}{5}}=\frac{4}{6}=\frac{2}{3}\)

Step 3: Take the square root

\(\tan\frac{x}{2}=\sqrt{\frac{2}{3}}\), rationalize the denominator:
\(\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{\sqrt{6}}{3}\)

We can also use the formula \(\tan\frac{x}{2}=\frac{1 - \cos x}{\sin x}\). Since \(\sin x=\sqrt{1-\cos^{2}x}=\sqrt{1 - (\frac{1}{5})^{2}}=\frac{2\sqrt{6}}{5}\) (we take the positive value of \(\sin x\) because if we consider the range of \(x\) such that \(\tan\frac{x}{2}\) is positive, and we can assume \(x\) is in a range where \(\sin x\) is positive for the positive value of \(\tan\frac{x}{2}\)). Then \(1-\cos x=1-\frac{1}{5}=\frac{4}{5}\), so \(\tan\frac{x}{2}=\frac{\frac{4}{5}}{\frac{2\sqrt{6}}{5}}=\frac{4}{2\sqrt{6}}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}\) (after rationalizing the denominator).

Answer:

\(\frac{\sqrt{6}}{3}\)