# Given sin(θ) = (sqrt(5 - sqrt(5)))/5 and cos(θ) > 0 how do you use identities to find the exact functional value of tan theta?

Jun 17, 2015

From ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$ and $\cos \theta > 0$ we get:

$\cos \theta = \sqrt{1 - {\sin}^{2} \theta} = \frac{\sqrt{20 + \sqrt{5}}}{5}$

$\tan \theta = \sin \frac{\theta}{\cos} \theta =$

#### Explanation:

${\sin}^{2} \alpha + {\cos}^{2} \alpha = 1$ for all $\alpha \in \mathbb{R}$

We are given $\cos \theta > 0$

So:

$\cos \theta = \sqrt{1 - {\sin}^{2} \theta}$

$= \sqrt{1 - {\left(\frac{\sqrt{5 - \sqrt{5}}}{5}\right)}^{2}}$

$= \sqrt{1 - \frac{5 - \sqrt{5}}{25}}$

$= \sqrt{\frac{25 - \left(5 - \sqrt{5}\right)}{25}}$

$= \frac{\sqrt{20 + \sqrt{5}}}{5}$

$\tan \theta = \sin \frac{\theta}{\cos} \theta$

$= \frac{\frac{\sqrt{5 - \sqrt{5}}}{5}}{\frac{\sqrt{20 + \sqrt{5}}}{5}}$

$= \frac{\sqrt{5 - \sqrt{5}}}{\sqrt{20 + \sqrt{5}}}$

$= \sqrt{\frac{5 - \sqrt{5}}{20 + \sqrt{5}}}$

$= \sqrt{\frac{5 - \sqrt{5}}{20 + \sqrt{5}} \cdot \frac{20 - \sqrt{5}}{20 - \sqrt{5}}}$

$= \sqrt{\frac{105 - 25 \sqrt{5}}{395}}$

$= \sqrt{\frac{21 - 5 \sqrt{5}}{79}}$