Question

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

Explanation:

Step1: Recall the half - angle formula for tangent

The half - angle formula for tangent is \(\tan\frac{\theta}{2}=\sqrt{\frac{1 - \cos\theta}{1+\cos\theta}}\) (we take the positive root since we are looking for the positive value of \(\tan\frac{1}{2}A\)).

Step2: Substitute \(\cos A=\frac{1}{2}\) into the formula

Substitute \(\cos\theta=\cos A = \frac{1}{2}\) into the formula \(\tan\frac{\theta}{2}=\sqrt{\frac{1 - \cos\theta}{1+\cos\theta}}\).
First, calculate the numerator: \(1-\cos A=1 - \frac{1}{2}=\frac{1}{2}\)
Then, calculate the denominator: \(1 + \cos A=1+\frac{1}{2}=\frac{3}{2}\)

Step3: Simplify the fraction inside the square root

\(\frac{1 - \cos A}{1+\cos A}=\frac{\frac{1}{2}}{\frac{3}{2}}\)
When dividing by a fraction, we multiply by its reciprocal, so \(\frac{\frac{1}{2}}{\frac{3}{2}}=\frac{1}{2}\times\frac{2}{3}=\frac{1}{3}\)

Step4: Take the square root

\(\tan\frac{1}{2}A=\sqrt{\frac{1}{3}}\)
Rationalize the denominator: \(\sqrt{\frac{1}{3}}=\frac{\sqrt{1}}{\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\) (multiply numerator and denominator by \(\sqrt{3}\))

Answer:

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