8. Triangular & Circular Functions

Lesson

We can find the trigonometric ratio for any angle on the $xy$`x``y`-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`=√`x`2+`y`2.

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 |

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°?

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.

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 $\sin\theta$

`s``i``n``θ`.Find $\cos\theta$

`c``o``s``θ`.Find $\tan\theta$

`t``a``n``θ`.Find $\csc\left(\theta\right)$

`c``s``c`(`θ`).Find $\sec\left(\theta\right)$

`s``e``c`(`θ`).Find $\cot\left(\theta\right)$

`c``o``t`(`θ`).

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 $\sin\theta$

`s``i``n``θ`, $\cos\theta$`c``o``s``θ`and $\tan\theta$`t``a``n``θ`.$\sin\theta=\editable{}$

`s``i``n``θ`=$\cos\theta=\editable{}$

`c``o``s``θ`=$\tan\theta=\editable{}$

`t``a``n``θ`=Hence or otherwise find the values of $\sec\theta$

`s``e``c``θ`, $\csc\theta$`c``s``c``θ`and $\cot\theta$`c``o``t``θ`.$\sec\theta=\editable{}$

`s``e``c``θ`=$\operatorname{cosec}\theta=\editable{}$

`c``o``s``e``c``θ`=$\cot\theta=\editable{}$

`c``o``t``θ`=

We just determined that for a point $P$`P`$\left(x,y\right)$(`x`,`y`) on the $xy$`x``y`-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.

$\sin\theta=\frac{y}{r}$`s``i``n``θ`=`y``r`, but $r$`r` is a length and is always positive, so this means that $\sin\theta$`s``i``n``θ` 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)$(`I``I`).

Sine Positive, Sine Negative |

$\cos\theta=\frac{x}{r}$`c``o``s``θ`=`x``r`, but $r$`r` is a length and is always positive, so this means that $\cos\theta$`c``o``s``θ` 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)$(`I``V`).

Cosine Positive, Cosine Negative |

$\tan\theta=\frac{y}{x}$`t``a``n``θ`=`y``x`, so this means that $\tan\theta$`t``a``n``θ` 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)$(`I``I``I`).

Tangent Positive, Tangent Negative |

If $\sec\theta=-\sqrt{2}$`s``e``c``θ`=−√2 and $\tan\theta=-1$`t``a``n``θ`=−1, in which quadrant is $\theta$`θ`?

**Think:** We have $\sec\theta=\frac{r}{x}$`s``e``c``θ`=`r``x` negative and $\tan\theta=\frac{y}{x}$`t``a``n``θ`=`y``x` negative.

**Do:** Therefore, if $\sec\theta$`s``e``c``θ` is negative it must be in quadrant $II$`I``I` or $III$`I``I``I` where $x$`x` is negative. If $\tan\theta$`t``a``n``θ` is negative in must in quadrant $II$`I``I` or $IV$`I``V` where $x$`x` and $y$`y` have different signs. For both of these to be true it must be in quadrant $II$`I``I`.

If $\theta$`θ` is an angle such that $\sin\theta$`s``i``n``θ`$>$>$0$0 and $\cos\theta$`c``o``s``θ`$<$<$0$0, which quadrant(s) does it lie in?

quadrant $I$

`I`Aquadrant $II$

`I``I`Bquadrant $IV$

`I``V`Cquadrant $III$

`I``I``I`Dquadrant $I$

`I`Aquadrant $II$

`I``I`Bquadrant $IV$

`I``V`Cquadrant $III$

`I``I``I`D

If $\theta$`θ` is an angle such that $\tan\theta$`t``a``n``θ`$>$>$0$0 and $\cot\left(\theta\right)$`c``o``t`(`θ`)$>$>$0$0, which quadrant(s) does it lie in?

quadrant $IV$

`I``V`Aquadrant $III$

`I``I``I`Bquadrant $II$

`I``I`Cquadrant $I$

`I`Dquadrant $IV$

`I``V`Aquadrant $III$

`I``I``I`Bquadrant $II$

`I``I`Cquadrant $I$

`I`D

If $P$`P`$\left(x,y\right)$(`x`,`y`) corresponds to a point on the unit circle at a rotation of $\left(\theta\right)$(`θ`) such that $\sin\theta=\frac{3}{5}$`s``i``n``θ`=35 and $\cos\theta=-\frac{4}{5}$`c``o``s``θ`=−45, which quadrant is $\left(x,y\right)$(`x`,`y`) located in?

quadrant $II$

`I``I`Aquadrant $IV$

`I``V`Bquadrant $III$

`I``I``I`Cquadrant $I$

`I`Dquadrant $II$

`I``I`Aquadrant $IV$

`I``V`Bquadrant $III$

`I``I``I`Cquadrant $I$

`I`D

Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle.