8.05 Trigonometric functions on the unit circle

Using a point on the unit circle

Previously, we have looked at the idea that if we plot a point on the $xy$xy-plane, $P$P$\left(x,y\right)$(x,y), we can use the coordinates of that point in conjunction with distance to the origin, $r$r, to find the trigonometric ratios for the angle formed with the positive $x$x-axis.

The unit circle for $\sin\theta$sinθ and $\cos\theta$cosθ

Instead of any point, let's look at points on a circle of radius $1$1, centered at the origin. Consider a point $P$P with coordinates $\left(1,0\right)$(1,0). This point lies on the circle, sitting on the $x$x-axis $1$1 unit to the right of the origin. As $P$P moves along the circle, how do its coordinates change?

Exploration

Picture a string of unit length, fixed to the origin at one end and tied to the point $P$P at the other. Initially the string lies along the $x$x-axis, so that $P$P is at $\left(1,0\right)$(1,0). By keeping the string taught, we can move $P$P around in a circle by changing the angle that the string makes with the $x$x-axis. Let’s call the angle $\theta$θ, and have positive values of $\theta$θ correspond to counterclockwise rotations about the origin, and negative values of $\theta$θ correspond to clockwise rotations about the origin.

We’ll focus our investigation for the time being on the first quadrant, where the coordinate values of $x$x and $y$y are both positive and $0^\circ\le\theta<90^\circ$0°≤θ<90°. For any value of $\theta$θ in this interval, we can construct the following right triangle.

Notice in this triangle that the hypotenuse has length $1$1. We can also relate the shorter side lengths to the coordinates of $P$P: the horizontal side has length $x$x, and the vertical side has length $y$y.

Now we can apply the main trigonometric ratios to the sides of this triangle to obtain the following relation:

$\cos\theta=\frac{x}{1}=x$cosθ=x1​=x
$\sin\theta=\frac{y}{1}=y$sinθ=y1​=y

So the coordinates of $P$P are $\left(x,y\right)=\left(\cos\theta,\sin\theta\right)$(x,y)=(cosθ,sinθ). Let’s test this on the boundaries of the first quadrant. When $\theta=0^\circ$θ=0°, we expect $P$P to be on the $x$x-axis, and indeed we see that $\cos0^\circ=1$cos0°=1 and $\sin0^\circ=0$sin0°=0, so that $\left(\cos0^\circ,\sin0^\circ\right)=\left(1,0\right)$(cos0°,sin0°)=(1,0). Now when $\theta=90^\circ$θ=90° we get $\cos90^\circ=0$cos90°=0 and $\sin90^\circ=1$sin90°=1, so that $\left(\cos90^\circ,\sin90^\circ\right)=\left(0,1\right)$(cos90°,sin90°)=(0,1), which is the point one unit above the origin.

 

 

 

Exact values

In general the value of the $x$x- and $y$y-coordinate of $P$P will be numbers that cannot be expressed as simple ratios of integers or radicals. But there are three values of $\theta$θ within each quadrant that do have familiar coordinates that are easy to work with. These values relate back to the exact value triangles, shown below.

If we scale the side lengths of these triangles so that each hypotenuse has length $1$1, then we obtain the following corresponding exact value triangles in the unit circle.

            

         

Worked example

Question 1

Use the unit circle to find the value of $\sec60^\circ$sec60°.

Think: If $P=\left(x,y\right)$P=(x,y) is a point on the unit circle, at an angle of $\theta$θ from the $x$x-axis, then $\sec\theta=\frac{1}{x}$secθ=1x​. We can also think of this as the ratio of the hypotenuse (which always has length $1$1) and the side adjacent the angle $\theta$θ (which always has length $x$x).

Do: The relevant exact value triangle is shown below.

$\sec\theta$secθ	$=$=	$\frac{\text{hypotenuse }}{\text{adjacent }}$hypotenuse adjacent ​	Definition of $\sec\theta$secθ
	$=$=	$\frac{1}{x}$1x​	Based on our findings from the unit circle exploration
	$=$=	$\frac{1}{\frac{1}{2}}$112​​	Using the diagram above
	$=$=	$2$2	Simplifying

 

Practice questions

Question 2

Question 3

 

The unit circle in all four quadrants

As we saw above, we can use a point on the unit circle to determine any of the six trigonometric ratios. We want to now consolidate the patterns and relationships which we can see in the unit circle.

Exploration

Have a look at the following applet, you can make settings for degrees or radians, and whether you want to rotate positively (counterclockwise) or negatively (clockwise). As you explore, see if you can determine where sine, cosine and tangent are positive or negative.

Guiding questions

1. In which quadrants are each of sine, cosine and tangent positive?
2. For each quadrant, list out the ratios which are positive there.
3. If we know where sine, cosine and tangent are positive/negative, what does this tell us about cosecant, secant and cotangent?
4. What do you notice about angles with the same reference angle? Such as $\sin\frac{\pi}{6}$sinπ6​, $\sin\frac{5\pi}{6}$sin5π6​, $\sin\frac{7\pi}{6}$sin7π6​ and $\sin\frac{11\pi}{6}$sin11π6​.

We should find that:

• The sine function is positive in the first and second quadrants and negative in third and fourth
• The cosine function is positive in the first and fourth quadrants and negative in second and third
• The tangent function is positive in the first and third quadrants and negative in second and fourth

These can be remembered by having a mental picture of the unit circle diagram or by means of the mnemonic ASTC: 'All Stations To Central: All-Sine-Tangent-Cosine' that shows which functions are positive in each quadrant. 

 

Reference angles

Recall that a reference angle is the acute angle which is formed with the closest $x$x-axis. For an angle, $\theta$θ, we typically use $\alpha$α for the reference angle.

Using what we know about the sign of the ratios in a particular quadrant, we can determine the sign of a ratio, then we can use the reference angle to determine the value of the ratio.

 

Worked examples

Question 4

By reference to the diagram below, give the values of each of the six trigonometric functions when $\theta=120^{\circ}$θ=120∘.

Think: We have the coordinates, $x=\frac{-1}{2}$x=−12​ and $y=\frac{\sqrt{3}}{2}$y=√32​, so we just need to use the definitions from the unit circle.

Do: From the coordinates we have:

$\cos120^{\circ}=-\frac{1}{2}$cos120∘=−12​ and $\sin120^{\circ}=\frac{\sqrt{3}}{2}$sin120∘=√32​

From these we find:

$\tan120^{\circ}=-\sqrt{3}$tan120∘=−√3 , $\sec120^{\circ}=-2$sec120∘=−2,  $\csc120^{\circ}=\frac{2}{\sqrt{3}}$csc120∘=2√3​, and $\cot120^{\circ}=-\frac{1}{\sqrt{3}}$cot120∘=−1√3​.

 

Practice questions

QUESTION 5

Question 6