Chapter 13.8, Problem 6WE

To determine

To find: The trigonometric function in a single angle.

Answer to Problem 6WE

The trigonometric function of a single angle is $\cos \frac{\alpha }{2}$ .

Explanation of Solution

Given information:

The function is $\sqrt{\frac{1+\cos \alpha }{2}},0°<\alpha <180°$

Calculation:

Consider the formula.

  $\cos \left(\alpha +\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta$

Thus,

  $\sqrt{\frac{1+\cos \alpha }{2}}=\sqrt{\frac{1+\cos \left(\frac{\alpha }{2}+\frac{\alpha }{2}\right)}{2}}$

Apply the formula.

  $\begin{array}{c}\sqrt{\frac{1+\cos \left(\frac{\alpha }{2}+\frac{\alpha }{2}\right)}{2}}=\sqrt{\frac{1+\cos \frac{\alpha }{2}\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2}\sin \frac{\alpha }{2}}{2}}\\ =\sqrt{\frac{1+{\cos }^2\frac{\alpha }{2}-{\sin }^2\frac{\alpha }{2}}{2}}\\ =\sqrt{\frac{{\cos }^2\frac{\alpha }{2}+{\cos }^2\frac{\alpha }{2}}{2}}\\ =\sqrt{\frac{2{\cos }^2\frac{\alpha }{2}}{2}}\\ =\sqrt{{\cos }^2\frac{\alpha }{2}}\\ =\cos \frac{\alpha }{2}\end{array}$