The terminal side of theta lies on the line 4x+3y=0 in quadrant IV, how do you find the values of the six trigonometric functions by finding a point on the line?

Feb 3, 2017

$\sin \theta = - \frac{4}{5}$; $\cos \theta = \frac{3}{5}$ ; $\tan \theta = - \frac{4}{3}$ ; $\sec \theta = \frac{5}{3}$; $\csc \theta = - \frac{5}{4}$ and $\cot \theta = - \frac{3}{4}$

Explanation:

The slope of the given line is ${\tan}^{- 1} \left(\frac{- 4}{3}\right)$ or $a r c \tan \left(\frac{- 4}{3}\right)$. This means there is a point on this line with coordinates (3, -4). Its radial distance from the origin would be $\sqrt{{3}^{3} + {4}^{2}}$ =5

The angle $\theta$ is, therefore, such that $\tan \theta = \frac{- 4}{3}$.

As shown in the figure below, triangle OPM is a rt. triangle, its hypotenuse is 5 with base and altitude being 3 and -4 respectively. Accordingly,
$\sin \theta = - \frac{4}{5}$; $\cos \theta = \frac{3}{5}$ ; $\tan \theta = - \frac{4}{3}$ ; $\sec \theta = \frac{5}{3}$; $\csc \theta = - \frac{5}{4}$ and $\cot \theta = - \frac{3}{4}$ 