# How do you use the unit circle to find values of cscx, secx and cotx?

Feb 13, 2015

Start from the definitions:

$\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$; $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$;

$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$; $\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$

Based on this, all we need to define using the unit circle are $\sin \left(x\right)$ and $\cos \left(x\right)$.

By definition, $\sin \left(x\right)$ is an ordinate (Y-coordinate) and $\cos \left(x\right)$ is an abscissa (X-coordinate) of a point lying on a unit circle at the end of a radius that forms an angle $x$ radians with the positive direction of the X-axis (counterclockwise from X-axis to this radius).

Using all the above, let's, for example, find $\sec \left(5 \frac{\pi}{6}\right)$.
$\sec \left(5 \frac{\pi}{6}\right) = \frac{1}{\cos} \left(5 \frac{\pi}{6}\right)$

Angle $5 \frac{\pi}{6} = {150}^{0}$ in a unit circle is determined by a radius from an origin of coordinates $O$ to a point $A$ in the second quadrant such that an angle ∠XOA=5pi/6. Drop a perpendicular from point $A$ on the X-axis. Its base, point $B$, has a coordinate $- \frac{\sqrt{3}}{2}$. This is obvious from the triangle ΔOAB. We can conclude that abscissa of point $A$ equals to $- \frac{\sqrt{3}}{2}$.

Therefore,
$\cos \left(5 \frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$
From this we find
$\sec \left(5 \frac{\pi}{6}\right) = - \frac{2}{\sqrt{3}} = - 2 \frac{\sqrt{3}}{3}$