# What is the equation of the tangent line of r=sintheta + costheta at theta=pi/2?

Sep 4, 2016

For the inward normal, $\theta = \pi$. For the outward normal in the opposite direction, it is $\theta = \frac{3}{2} \pi$, obtained by adding $\pi$ to $\theta$.

#### Explanation:

Here,#

$r = \sin \theta + \cos \theta$

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

$= \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) \cos \theta + \sin \left(\frac{\pi}{4}\right) \sin \theta\right)$

$= \sqrt{2} \cos \left(\theta - \frac{\pi}{4}\right)$

This represents the circle with center at the pole r = 0 and diameter

sqrt2. The radius is $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. The initial line is along a

diameter , $\theta = \frac{\pi}{4}$.

The normal at a point on the circle is along the radius to the point.

Thus, the radial line $\theta = \frac{\pi}{2}$ is outward normal to the point

under reference, $\left(\sqrt{2} , \frac{\pi}{2}\right)$.

For the inward normal, the equation is

$\theta = \frac{3}{2} \pi$, for the opposite direction.