Question

use the adjacent figure to find the exact value of the following trigonometric function. $\cos \frac { \alpha } { 2 }$

Explanation:

Step1: Find the hypotenuse

By Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\), where \(a = 3\), \(b=4\).
\(c=\sqrt{3^{2}+4^{2}}=\sqrt{9 + 16}=\sqrt{25}=5\)
So, \(\cos\alpha=\frac{3}{5}\)

Step2: Use the half - angle formula

The half - angle formula for cosine is \(\cos\frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}}\) (since \(\alpha\) is an acute angle in a right - triangle, \(\frac{\alpha}{2}\) is also acute and \(\cos\frac{\alpha}{2}>0\))
Substitute \(\theta=\alpha\) and \(\cos\alpha=\frac{3}{5}\) into the formula:
\(\cos\frac{\alpha}{2}=\sqrt{\frac{1+\frac{3}{5}}{2}}=\sqrt{\frac{\frac{5 + 3}{5}}{2}}=\sqrt{\frac{\frac{8}{5}}{2}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\)

Answer:

\(\frac{2\sqrt{5}}{5}\)