Question

If $z=\tan \frac{\alpha }{2}$, show that localid="1646416257885" $\cos \alpha =\frac{1-z^2}{1+z^2}$.

Short answer

To show that $\cos \alpha =\frac{1-z^2}{1+z^2}$,first use double angle and inverse trigonometry, and simplify the equation.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} z=\tan \frac{\alpha }{2}, \\ \cos \alpha =\frac{1-z^2}{1+z^2} \end{gathered}$$

Using the double-angle formula,

$\cos \left(2\theta \right)=2{\cos }^2\theta -1$

From inverse trigonometry,

$\cos \left({\cos }^{-1}x\right)=x$

Then,

${\tan }^{-1}={\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$

Step 2. Simplify the equation.

Consider the given question,

$$\begin{gathered} z=\tan \frac{\alpha }{2} \\ \alpha =2{\tan }^{-1}z \\ \alpha =2{\cos }^{-1}\left(\frac{1}{\sqrt{1+z^2}}\right) \end{gathered}$$

Using the double-angle formula and inverse trigonometry,

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