5.04 Trigonometric functions in the coordinate plane

Determining trigonometric ratios from a point

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.

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.

1. What do you notice about the signs of the ratios if you were to simplify them?
2. 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 2

Question 3

Identifying quadrants

We just determined that for a point $P$P$\left(x,y\right)$(x,y) on the $xy$xy-plane which is $r$r units from the origin, we can create an angle $\theta$θ with the positive $x$x-axis and the trigonometric ratios are defined as follows. 

$\sin\theta$sinθ	$=$=	$\frac{y}{r}$yr​
$\cos\theta$cosθ	$=$=	$\frac{x}{r}$xr​
$\tan\theta$tanθ	$=$=	$\frac{y}{x}$yx​
$\csc\theta$cscθ	$=$=	$\frac{r}{y}$ry​
$\sec\theta$secθ	$=$=	$\frac{r}{x}$rx​
$\cot\theta$cotθ	$=$=	$\frac{x}{y}$xy​

 

Consider what this means about where the different ratios are positive and negative. We will just look at sine, cosine and tangent as the reciprocal ratios will have the same sign as the primary ratio.

Sine ratio

$\sin\theta=\frac{y}{r}$sinθ=yr​, but $r$r is a length and is always positive, so this means that $\sin\theta$sinθ will be positive where $y$y is positive. This will be in quadrants $1$1 $\left(I\right)$(I)and $2$2$\left(II\right)$(II).

Sine Positive, Sine Negative

Cosine ratio

$\cos\theta=\frac{x}{r}$cosθ=xr​, but $r$r is a length and is always positive, so this means that $\cos\theta$cosθ will be positive where $x$x is positive. This will be in quadrants $1$1$\left(I\right)$(I) and $4$4$\left(IV\right)$(IV).

Cosine Positive, Cosine Negative

Tangent ratio

$\tan\theta=\frac{y}{x}$tanθ=yx​, so this means that $\tan\theta$tanθ will be positive where $x$x and $y$y is are either both positive or both negative. This will be in quadrants $1$1$\left(1\right)$(1) and $3$3$\left(III\right)$(III).

Tangent Positive, Tangent Negative

 

Worked examples

Question 4

If $\sec\theta=-\sqrt{2}$secθ=−√2 and $\tan\theta=-1$tanθ=−1, in which quadrant is $\theta$θ?

Think: We have $\sec\theta=\frac{r}{x}$secθ=rx​ negative and $\tan\theta=\frac{y}{x}$tanθ=yx​ negative. 

Do: Therefore, if $\sec\theta$secθ is negative it must be in quadrant $II$II or $III$III where $x$x is negative. If $\tan\theta$tanθ is negative in must in quadrant $II$II or $IV$IV where $x$x and $y$y have different signs. For both of these to be true it must be in quadrant $II$II.

 

Practice questions

QUESTION 5

QUESTION 6

QUESTION 7