# Question #cbbe5

Jul 16, 2017

Simplifying it first.

#### Explanation:

$r = 10 \csc \left(\theta + \frac{7 \pi}{4}\right)$
Multiply by sine.
$r \sin \left(\theta + \frac{7 \pi}{4}\right) = 10$
Now use the sine addition formula:
$\sin \left(\theta + \frac{7 \pi}{4}\right) = \sin \left(\theta\right) \cos \left(\frac{7 \pi}{4}\right) + \cos \left(\theta\right) \sin \left(\frac{7 \pi}{4}\right)$
Since $\frac{7 \pi}{4}$ is in quadrant IV, sine is negative and cosine is positive, with
$\sin \left(\frac{7 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$
and
$\cos \left(\frac{7 \pi}{4}\right) = \frac{\sqrt{2}}{2}$

Therefore we have...
$r \sin \left(\theta + \frac{7 \pi}{4}\right) = 10$
$r \sin \left(\theta\right) \cos \left(\frac{7 \pi}{4}\right) + r \cos \left(\theta\right) \sin \left(\frac{7 \pi}{4}\right) = 10$
$- r \sin \left(\theta\right) \left(\frac{\sqrt{2}}{2}\right) + r \cos \left(\theta\right) \left(\frac{\sqrt{2}}{2}\right) = 10$
Multiply both sides by $\sqrt{2}$. This has the same effect as dividing by $\frac{\sqrt{2}}{2}$.
$- r \sin \left(\theta\right) + r \cos \left(\theta\right) = 10 \sqrt{2}$
Now use the standard polar substitutions:
$x = r \cos \left(\theta\right) \mathmr{and} y = r \sin \left(\theta\right)$.
We have...
$- y + x = 10 \sqrt{2}$
Solve for y, or put it in whatever form you desire. This is a line.