# How do you determine costheta given tantheta=2/3, pi<theta<(3pi)/2?

Dec 17, 2016

#### Explanation:

Start with the identity ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$

Divide both sides by ${\cos}^{2} \left(\theta\right)$:

${\cos}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) + {\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) = \frac{1}{\cos} ^ 2 \left(\theta\right)$

Use the identity $\sin \frac{\theta}{\cos} \left(\theta\right) = \tan \left(\theta\right)$ and ${\cos}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) = 1$:

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

Solve for $\cos \left(\theta\right)$:

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

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

The domain restriction $\pi < \theta < \frac{3 \pi}{2}$ tells us that $\theta$ is in the third quadrant (where the cosine function is negative), therefore, we change the $\pm$ to - only:

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

Substitute ${\left(\frac{2}{3}\right)}^{2}$ for ${\tan}^{2} \left(\theta\right)$

$\cos \left(\theta\right) = - \sqrt{\frac{1}{1 + {\left(\frac{2}{3}\right)}^{2}}}$

$\cos \left(\theta\right) = - \sqrt{\frac{1}{1 + \frac{4}{9}}}$

$\cos \left(\theta\right) = - \sqrt{\frac{1}{\frac{13}{9}}}$

$\cos \left(\theta\right) = - \sqrt{\frac{9}{13}}$

$\cos \left(\theta\right) = - \frac{3 \sqrt{13}}{13}$