We can find the trigonometric ratio for any angle on the $xy$xy-plane by using the coordinates of a point on the terminal side of the angle. The image below shows a points $P$P $\left(x,y\right)$(x,y), with a terminal side length $r$r.
Angle $\theta$θ with terminal side end point of $\left(x,y\right)$(x,y)
For the point $P$P$\left(x,y\right)$(x,y), we can see that this forms a triangle with hypotenuse of length $r$r, opposite side of length $y$y and adjacent side of length $x$x. Using the Pythagorean theorem, we can see that $r=\sqrt{x^2+y^2}$r=√x2+y2.
To be more general, let's consider the point $P$P$\left(x,y\right)$(x,y), terminal side length $r$r, we get the following:
| $\sin\theta$sinθ | $=$= | $\frac{\text{opposite }}{\text{hypotenuse }}$opposite hypotenuse | $=$= | $\frac{y}{r}$yr |
| $\cos\theta$cosθ | $=$= | $\frac{\text{adjacent }}{\text{hypotenuse }}$adjacent hypotenuse | $=$= | $\frac{x}{r}$xr |
| $\tan\theta$tanθ | $=$= | $\frac{\text{opposite }}{\text{adjacent }}$opposite adjacent | $=$= | $\frac{y}{x}$yx |
| $\csc\theta$cscθ | $=$= | $\frac{\text{hypotenuse }}{\text{opposite }}$hypotenuse opposite | $=$= | $\frac{r}{y}$ry |
| $\sec\theta$secθ | $=$= | $\frac{\text{hypotenuse }}{\text{opposite }}$hypotenuse opposite | $=$= | $\frac{r}{x}$rx |
| $\cot\theta$cotθ | $=$= | $\frac{\text{adjacent }}{\text{opposite }}$adjacent opposite | $=$= | $\frac{x}{y}$xy |
Exploration
Use this GeoGebra widget to explore what happens as the terminal side moves through different quadrants.
- What do you notice about the signs of the ratios if you were to simplify them?
- What do you notice about ratios with the same reference angle such as $30^\circ$30°, $150^\circ$150°, $210^\circ$210° and $330^\circ$330°?
Worked example
Question 1
For the angle, $\theta$θ, formed by the positive $x$x-axis and terminal side with endpoint $P$P$\left(3,-4\right)$(3,−4), Complete the table below.
| $\sin\theta$sinθ | $\cos\theta$cosθ | $\tan\theta$tanθ | $\csc\theta$cscθ | $\sec\theta$secθ | $\cot\theta$cotθ |
Think: We have $x=3$x=3, $y=-4$y=−4. We first need to find $r$r and then we just use our ratios.
Do:
| $r$r | $=$= | $\sqrt{x^2+y^2}$√x2+y2 |
| $=$= | $\sqrt{3^2+\left(-4\right)^2}$√32+(−4)2 | |
| $=$= | $\sqrt{9+16}$√9+16 | |
| $=$= | $\sqrt{25}$√25 | |
| $=$= | $5$5 |
Now we have $x=3$x=3 which is like our adjacent side, $y=-4$y=−4 which is like our opposite side and $r=5$r=5 which is like our hypotenuse.
| $\sin\theta$sinθ | $\cos\theta$cosθ | $\tan\theta$tanθ | $\csc\theta$cscθ | $\sec\theta$secθ | $\cot\theta$cotθ |
| $-\frac{4}{5}$−45 | $\frac{3}{5}$35 | $\frac{-4}{3}$−43 | $-\frac{5}{4}$−54 | $\frac{5}{3}$53 | $\frac{-3}{4}$−34 |
Reflect: This point would be in quadrant $4$4, and only cosine and secant are positive. Depending on the quadrant, different ratios will have different signs.
Practice questions
Question 1
The point on the graph has coordinates $\left(15,8\right)$(15,8).
Find $r$r, the distance from the point to the origin.
Find $\sin\theta$sinθ.
Find $\cos\theta$cosθ.
Find $\tan\theta$tanθ.
Find $\csc\left(\theta\right)$csc(θ).
Find $\sec\left(\theta\right)$sec(θ).
Find $\cot\left(\theta\right)$cot(θ).
Question 2
The point on the graph has coordinates $\left(7,24\right)$(7,24).
Find $r$r, the distance from the point to the origin.
Consider the diagram below. What are the coordinates of this point?
Considering the diagram above, or otherwise, find the values of $\sin\theta$sinθ, $\cos\theta$cosθ and $\tan\theta$tanθ.
$\sin\theta=\editable{}$sinθ=
$\cos\theta=\editable{}$cosθ=
$\tan\theta=\editable{}$tanθ=
Hence or otherwise find the values of $\sec\theta$secθ, $\csc\theta$cscθ and $\cot\theta$cotθ.
$\sec\theta=\editable{}$secθ=
$\csc\theta=\editable{}$cscθ=
$\cot\theta=\editable{}$cotθ=
Question 3
The point on the graph has coordinates $\left(-7,-24\right)$(−7,−24).
Find $r$r, the distance from the point to the origin.
Find $\sin\theta$sinθ.
Find $\cos\theta$cosθ.
Find $\tan\theta$tanθ.
Find $\csc\left(\theta\right)$csc(θ).
Find $\sec\left(\theta\right)$sec(θ).
Find $\cot\left(\theta\right)$cot(θ).