Question

A terminal ray passes through (12,-5). Fine the value to all six trig functions

Answer 1

Okay, here we have this:

Considering the provided angle, we are going to calculate the requested trigonometric functions, so we obtain the following:

So the first thing we will do is calculate the length of the hypotenuse, that is, the distance between the given point and the origin, then we have:

$\begin{gathered} r=\sqrt{(12-0)^2+(-5-0)^2} \\ r=\sqrt{12^2+(-5)^2} \\ r=\sqrt{144+25} \\ r=\sqrt{169} \\ r=13 \end{gathered}$

Now we proceed to find the value of each ratio:

$sin\beta=\frac{y}{r}=\frac{-5}{13}$$cos\beta=\frac{x}{r}=\frac{12}{13}$$tan\beta=\frac{y}{x}=\frac{-5}{12}$$csc\beta=\frac{r}{y}=\frac{13}{-5}=-\frac{13}{5}$$sec\beta=\frac{r}{x}=\frac{13}{12}$$cot\beta=\frac{x}{y}=\frac{12}{-5}=-\frac{12}{5}$