5.03 The unit circle and angles of any magnitude

When we restricted our attention to right-angled triangles, we only considered the trigonometric ratios $\sin\theta$sinθ, $\cos\theta$cosθ $\tan\theta$tanθ for angles whose size was between $0^\circ$0° and $90^\circ$90°.

The unit circle

The unit circle is a visual tool that allows us to connect the trigonometric ratios to angles of any magnitude. It is defined as a circle with a radius of $1$1 unit centred at the origin on the Cartesian plane. A radius to any point $P$P on the circle is free to move and makes an angle $\theta$θ with the positive horizontal axis. By convention, the angle is measured anticlockwise from the positive horizontal axis. The angle can have any size, positive or negative, depending on how far the point has moved around the circle.

The sine of the angle is defined to be the $y$y-coordinate of point $P$P. For an acute angle $\theta$θ, this makes sense as we have always thought of:

$\sin\theta$sinθ

$=$=

$\frac{\text{opposite}}{\text{hypotenuse}}$oppositehypotenuse​

 

In the triangle above, this gives us:

$\sin\theta$sinθ	$=$=	$\frac{y\text{-coordinate of}\ P}{1}$y-coordinate of P1​
$y\text{-coordinate of}\ P$y-coordinate of P	$=$=	$\sin\theta$sinθ

 

The cosine of the angle is defined to be the $x$x-coordinate of point $P$P. For an acute angle $\theta$θ, this makes sense as we have always thought of:

$\cos\theta$cosθ

$=$=

$\frac{\text{adjacent}}{\text{hypotenuse}}$adjacenthypotenuse​

 

In the triangle above, this gives us:

$\cos\theta$cosθ	$=$=	$\frac{x\text{-coordinate of}\ P}{1}$x-coordinate of P1​
$x\text{-coordinate of}\ P$x-coordinate of P	$=$=	$\cos\theta$cosθ

 

The tangent of the angle can be geometrically defined to be $y$y-coordinate of point $Q$Q, where $Q$Q is the intersection of the extension of the line $OP$OP and the tangent of the circle at $\left(1,0\right)$(1,0). Using similar triangles we can define this algebraically as the ratio:

$\frac{\sin\theta}{\cos\theta}$sinθcosθ​

$=$=

$\frac{\tan\theta}{1}$tanθ1​

This also represents the gradient of the line that forms the angle $\theta$θ to the positive $x$x-axis and hence this gives us our first trigonometric identity:

$\tan\theta$tanθ

$=$=

$\frac{\sin\theta}{\cos\theta}$sinθcosθ​   provided $\cos\theta\ne0$cosθ≠0

 

Using the applet below, change the angle and take note of the sign (positive or negative) that each ratio has in the four different quadrants. 

Created with Geogebra

Depending on where the point is on the unit circle, we say the angle is in one of four quadrants. From the applet, as we watch the point moving through the quadrants, it can be seen that the sine function is positive for angles in the first and second quadrants; cosine is positive for angles in the first and fourth quadrants; and consequently, the tangent is positive for angles in the first and third quadrants.

These can be remembered by having a mental picture of the unit circle diagram and the pattern ASTC: All-Sine-Tangent-Cosine which refers to which ratios are positive in each quadrant. You might have been taught the mnemonic "All Stations To Central" to help you remember this.

From the applet, we can also see that the values of $\sin\theta$sinθ and $\cos\theta$cosθ are bound between $1$1 and $-1$−1. However, the value of $\tan\theta$tanθ is not bound and is undefined at $90^\circ$90° and $270^\circ$270°. The pattern will restart at $360^\circ$360°. We will revisit this when we consider the graphs of the trigonometric functions later in this topic. 

Equivalent Angles

We can use symmetry within the circle to find equivalent angles. That is angles with the same trigonometric ratios.

From the diagram we can see that the $y$y-coordinate of point $A$A is the same as point $B$B, hence:

$\sin\theta$sinθ

$=$=

$\sin(180^\circ-\theta)$sin(180°−θ)

We can also see the $x$x-coordinate of $A$A is the same as point $D$D, hence:

$\cos\theta$cosθ

$=$=

$\sin(360^\circ-\theta)$sin(360°−θ)

The $y$y-coordinate of $C$C has the same magnitude but would be negative in comparison to the $x$x-coordinate of $A$A, hence:

$\sin(180+\theta)$sin(180+θ)

$=$=

$-\sin\theta$−sinθ

The trigonometric functions of all angles are defined in a manner that guarantees that a function of any angle will be related to the same function of an angle in the first quadrant. This first quadrant angle may be called the relative acute angle, or the reference angle. The relative acute angle is between $0^\circ$0° and $90^\circ$90°.

We can summarise trigonometric ratios for angles of any magnitude as follows:

First quadrant Angle $\theta$θ $\sin\theta$sinθ is positive $\cos\theta$cosθ is positive $\tan\theta$tanθ is positive	Second quadrant Angle $180^\circ-\theta$180°−θ $\sin(180^\circ-\theta)=\sin\theta$sin(180°−θ)=sinθ $\cos(180^\circ-\theta)=-\cos\theta$cos(180°−θ)=−cosθ $\tan(180^\circ-\theta)=-\tan\theta$tan(180°−θ)=−tanθ
Third quadrant Angle $180^\circ+\theta$180°+θ $\sin(180^\circ+\theta)=-\sin\theta$sin(180°+θ)=−sinθ $\cos(180^\circ+\theta)=-\cos\theta$cos(180°+θ)=−cosθ $\tan(180^\circ+\theta)=\tan\theta$tan(180°+θ)=tanθ	Fourth quadrant Angle $360^\circ-\theta$360°−θ $\sin(360^\circ-\theta)=-\sin\theta$sin(360°−θ)=−sinθ $\cos(360^\circ-\theta)=\cos\theta$cos(360°−θ)=cosθ $\tan(360^\circ-\theta)=-\tan\theta$tan(360°−θ)=−tanθ

Worked examples

Example 1

Express $\cos117^\circ$cos117° in terms of a first quadrant angle.

Think: The angle $117^\circ$117° is between $90^\circ$90° and $180^\circ$180°, so it is in the second quadrant. The point representing $117^\circ$117° on the unit circle diagram, where the radius cuts the circle, must have a negative horizontal coordinate.

Do: Therefore, $\cos117^\circ$cos117° must be the same as $-\cos\left(180^\circ-117^\circ\right)=-\cos63^\circ$−cos(180°−117°)=−cos63°.

Example 2

Find the exact value of $\cos240^\circ$cos240°.

Think: The angle $240^\circ$240° is in the third quadrant, and remembering ASTC, we know that $\cos240^\circ$cos240° will be negative.

Do: We find its first quadrant reference angle by subtracting $180^\circ$180° to get $60^\circ$60°.

 

Of course, we remember our exact values for $30^\circ$30°, $45^\circ$45° and $60^\circ$60°! Thus, the exact value of $\cos240^\circ=-\cos60^\circ=$cos240°=−cos60°= $-\frac{1}{2}$−12​.

Angles outside of the domain $0^{\circ}\le\theta\le360^{\circ}$0∘≤θ≤360∘

What other equivalences can you see around the unit circle? What about angles larger than $360^\circ$360° or negative angles?

An angle simply specifies a rotation. A rotation of more than $360^\circ$360° means we have completed more than a full lap around the unit circle. We will land on a point $P$P that we had already passed over. A negative angle means we are rotating in the opposite direction, so we set off clockwise rather than anticlockwise, stopping in a position we are used to thinking about being related to a positive angle between $0^\circ$0° and $360^\circ$360°. 

If you think about it, there are literally an infinite number of angles that have the same trigonometric ratio! This behaviour is due to the periodic nature of the trigonometric functions - in basic terms, the values will repeat. So to find the acute reference angle, first, if necessary, add or subtract multiples of $360^\circ$360° to obtain an angle between $0^\circ$0° and $360^\circ$360°. Then decide what quadrant the angle is in as we did before.

Practice questions

Question 1

Question 2

Question 3

Evaluating exact trigonometric ratios

Most of the time in this topic, you will evaluate trigonometric functions for given angles or find the angle that makes a ratio true. The following style of question asks you to find one trigonometric function having been given another. Perhaps counterintuitively, we do not need to find the actual angle involved at any stage! We simply use our understanding of the unit circle to ensure our ratio is presented with the correct sign. We use Pythagoras' theorem to find the exact length of any missing sides in a related triangle. These sides are used as the values in the trigonometric ratio requested.

Worked Example

Example 3

If $\cos\theta=\frac{2}{5}$cosθ=25​ and $\sin\theta<0$sinθ<0 find $\tan\theta$tanθ.

Think: If the cos ratio is positive but the sin ratio is negative, this angle $\theta$θ must lie in the fourth quadrant. The image below shows us the scenario:

Do: To evaluate $\tan\theta$tanθ, we must find the missing side in the triangle. Using Pythagoras, this side is:

$\sqrt{5^2-2^2}=\sqrt{21}$√52−22=√21

Remembering that $\tan\theta$tanθ is negative in the fourth quadrant, our final answer is:

$\tan\theta=-\frac{\sqrt{21}}{2}$tanθ=−√212​

Practice questions

QUESTION 4

QUESTION 5