# How do you use special angles to find the ratios of csc 315 degrees?

Oct 17, 2015

${315}^{\circ} = {45}^{\circ}$ less than a full circle
$\textcolor{w h i t e}{\text{XXX}}$i.e. ${315}^{\circ} \equiv - {45}^{\circ}$
$\csc \left(- {45}^{\circ}\right) = \frac{1}{\sin} \left(- {45}^{\circ}\right) = - \frac{1}{\sin} \left({45}^{\circ}\right) = - \sqrt{2}$

#### Explanation:

Since $\left(- {45}^{\circ}\right)$ is in Quadrant IV, $\sin$ (and therefore $\csc$) is negative.

The ${45}^{\circ}$ angle is a no-right angle in a right-angled equilateral triangle. If the equal length arms of the triangle are of length 1, then the hypotenuse is of length $\sqrt{2}$ (Pythagorean Theorem).

$\sin = \left(\text{opposite")/("hypotenuse}\right)$