# Question #110e4

Apr 10, 2017

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

#### Explanation:

It is given that $\sin \left(\theta\right) = k$ and that $\theta$ is an obtuse angle ($\frac{\pi}{2} < \theta < \pi$). Thus, angle $\theta$ is in the second quadrant.

We can find $\cos \left(\theta\right)$ by using the property ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$. Then, $\cos \left(\theta\right) = \pm \sqrt{1 - {\sin}^{2} \left(\theta\right)} = \pm \sqrt{1 - {k}^{2}}$. Since $\theta$ is in the second quadrant and the cosine function returns negative values in the second quadrant, $\cos \left(\theta\right) = - \sqrt{1 - {k}^{2}}$.

We can find $\tan \left(\theta\right)$ by using the property $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right) = \frac{k}{-} \sqrt{1 - {k}^{2}} = - \frac{k}{\sqrt{1 - {k}^{2}}}$. This can also be rewritten as $- \frac{k \sqrt{1 - {k}^{2}}}{1 - {k}^{2}}$.

To find $\sin \left(\theta + \pi\right)$, use the unit circle. $\pi$ is half a circle, and adding it to an angle negates both its $x$ and $y$ coordinate. $\sin \left(\theta\right)$ is the value of the $y$ coordinate; therefore $\sin \left(\theta + \pi\right) = - \sin \left(\theta\right) = - k$.