# How do you find the exact value of the six trigonometric functions of the angle whose terminal side passes through (2/3, 5/2)?

Jan 11, 2018

$\cos \left(\theta\right) = \frac{4}{\sqrt{241}}$, $\sin \left(\theta\right) = \frac{15}{\sqrt{241}}$, and $\tan \left(\theta\right) = \frac{15}{4}$, $\sec \left(\theta\right) = \frac{\sqrt{241}}{4}$, $\csc \left(\theta\right) = \frac{\sqrt{241}}{15}$, and $\cot \left(\theta\right) = \frac{4}{15}$.

#### Explanation:

The points $\left(\frac{2}{3} , \frac{5}{2}\right)$ and $\left(4 , 15\right)$ fall on the same line and therefore the terminal side of the same angle, $\theta$, and the second point is easier to work with. (I basically multiplied through by 6 to scale the point up.)

Calculate $r = \sqrt{{4}^{2} + {15}^{2}} = \sqrt{16 + 225} = \sqrt{241}$.

Knowing $x$, $y$, and $r$ we can find:

$\cos \left(\theta\right) = \frac{x}{r}$, $\sin \left(\theta\right) = \frac{y}{r}$, and $\tan \left(\theta\right) = \frac{y}{x}$.

For this problem we have:

$\cos \left(\theta\right) = \frac{4}{\sqrt{241}}$, $\sin \left(\theta\right) = \frac{15}{\sqrt{241}}$, and $\tan \left(\theta\right) = \frac{15}{4}$.

The remaining functions are found by taking reciprocals:

$\sec \left(\theta\right) = \frac{\sqrt{241}}{4}$, $\csc \left(\theta\right) = \frac{\sqrt{241}}{15}$, and $\cot \left(\theta\right) = \frac{4}{15}$.