The unit circle, the six trigonometric ratios and special angles – Fundamentals of Trigonometry (Senior 4 Mathematics) | Ganzaa

Fundamentals of Trigonometry

Lessons

Senior 4 · Mathematics

Fundamentals of Trigonometry

5 lessons

Lesson 2 of 5

The unit circle, the six trigonometric ratios and special angles

25 min

Beyond the right-angled triangle

In S3 you defined $\sin$ , $\cos$ and $\tan$ as ratios of sides in a right-angled triangle, which only works for acute angles. But a point on a turning wheel, an alternating current or a swinging pendulum needs $\sin 150^\circ$ , $\cos(-60^\circ)$ and $\tan 300^\circ$ . The fix is to read the ratios from a circle of radius 1: every angle, of any size and either sign, has a terminal side that meets this circle exactly once.

Definition

Unit (trigonometric) circle: the circle with centre the origin $O$ and radius 1, oriented anticlockwise, with starting point $A (1, 0)$ .

The six trigonometric ratios: if the terminal side of an angle $\theta$ in standard position meets the unit circle at $M (x, y)$ , then

\cos\theta = x, \qquad \sin\theta = y, \qquad \tan\theta = \frac{y}{x}\ (x \ne 0),

\cot\theta = \frac{x}{y}\ (y \ne 0), \qquad \sec\theta = \frac{1}{x}\ (x \ne 0), \qquad \csc\theta = \frac{1}{y}\ (y \ne 0).

( $\csc$ is also written $\operatorname{cosec}$ .) For an acute angle these agree with S3: the right triangle with hypotenuse $O M = 1$ has adjacent side $x$ and opposite side $y$ .

Since $M$ lies on a circle of radius 1, $-1 \le \cos\theta \le 1$ and $-1 \le \sin\theta \le 1$ for every $\theta$ . The tangent is the height at which the extended terminal side crosses the tangent line $x = 1$ , which is where its name comes from.

If instead the terminal side passes through any point $P (x, y)$ at distance $r = \sqrt{x^2 + y^2}$ from $O$ , similar triangles give $\sin\theta = \frac{y}{r}$ , $\cos\theta = \frac{x}{r}$ and $\tan\theta = \frac{y}{x}$ .

Signs in the four quadrants

Quadrant	Angle (radians)	$\sin\theta, \csc\theta$	$\cos\theta, \sec\theta$	$\tan\theta, \cot\theta$
I	$0 \lt \theta \lt \frac{\pi}{2}$	+	+	+
II	$\frac{\pi}{2} \lt \theta \lt \pi$	+	−	−
III	$\pi \lt \theta \lt \frac{3\pi}{2}$	−	−	+
IV	$\frac{3\pi}{2} \lt \theta \lt 2\pi$	−	+	−

Going round from quadrant I: All ratios are positive, then only Sine, then only Tangent, then only Cosine (with their reciprocals): A, S, T, C.

Worked example 1: a point on the terminal side

The terminal side of $\theta$ passes through $P (- 3, 4)$ . Find the six trigonometric ratios of $\theta$ .

Solution $r = \sqrt{(-3)^2 + 4^2} = 5$ . So

\sin\theta = \frac{4}{5}, \quad \cos\theta = -\frac{3}{5}, \quad \tan\theta = -\frac{4}{3}, \quad \csc\theta = \frac{5}{4}, \quad \sec\theta = -\frac{5}{3}, \quad \cot\theta = -\frac{3}{4} .

Check with the signs table: $P$ is in quadrant II, where only $\sin$ and $\csc$ are positive. ✓

Worked example 2: one ratio gives the other five

Given $\sin\theta = \frac{5}{13}$ and $\frac{\pi}{2} \lt \theta \lt \pi$ , find the other five ratios.

Solution Take $y = 5$ and $r = 13$ ; then $x^{2} = 13^{2} - 5^{2} = 144$ , so $x = \pm 12$ . In quadrant II, $x \lt 0$ , so $x = - 12$ .

\cos\theta = -\frac{12}{13}, \quad \tan\theta = -\frac{5}{12}, \quad \csc\theta = \frac{13}{5}, \quad \sec\theta = -\frac{13}{12}, \quad \cot\theta = -\frac{12}{5} .

Quadrantal and special angles

At the quadrantal angles the point $M$ is $(1, 0)$ , $(0, 1)$ , $(- 1, 0)$ or $(0, - 1)$ , so the values can be read off directly. For $\frac{\pi}{6}$ , $\frac{\pi}{4}$ and $\frac{\pi}{3}$ , two triangles give exact values:

$\theta$	$0$	$\frac{\pi}{6}$ (30°)	$\frac{\pi}{4}$ (45°)	$\frac{\pi}{3}$ (60°)	$\frac{\pi}{2}$ (90°)	$π \pi$	$\frac{3\pi}{2}$
$\sin\theta$	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1	0	$- 1$
$\cos\theta$	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0	$- 1$	0
$\tan\theta$	0	$\frac{\sqrt{3}}{3}$	1	$\sqrt{3}$	undefined	0	undefined

For example, in the half equilateral triangle, $\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2}$ and $\tan 60^\circ = \frac{\sqrt{3}}{1} = \sqrt{3}$ ; in the half square, $\sin 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .

Worked example 3: exact values

Find the exact value of (a) $2\sin\frac{\pi}{6}\cos\frac{\pi}{6}$ (b) $\tan^2\frac{\pi}{3} - \sec^2\frac{\pi}{4}$ (c) $\sin\frac{3\pi}{2} + \cos\pi \cdot \csc\frac{\pi}{6}$ .

Solution

$2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$ .
$\tan^2\frac{\pi}{3} = 3$ and $\sec\frac{\pi}{4} = \frac{1}{\cos\frac{\pi}{4}} = \sqrt{2}$ , so $3 - 2 = 1$ .
$- 1 + (- 1) (2) = - 3$ .

Warning

Some ratios do not exist. $\tan\frac{\pi}{2}$ would need $\frac{1}{0}$ : it is undefined, not 0 and not "infinity". Likewise $\cot 0$ , $\csc \pi$ and $\sec\frac{3\pi}{2}$ are undefined. A calculator shows "Math ERROR" for $\tan 90^\circ$ for this reason.

Tip

A pattern for the sines of $0, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ : $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$ . The cosines are the same list in reverse order.

Schattman Math: All 6 Trig Functions on the Unit Circle. Watch for where each of the six ratios appears as a length in one picture of the unit circle.

✦

Practice

Give exact values.

The terminal side of $\theta$ passes through $P (5, - 12)$ . Find the six ratios.
In which quadrant is $\theta$ if $\sin\theta \lt 0$ and $\tan\theta \gt 0$ ?
Given $\cos\theta = -\frac{3}{5}$ with $\pi \lt \theta \lt \frac{3\pi}{2}$ , find $\sin\theta$ , $\tan\theta$ and $\csc\theta$ .
Evaluate: (a) $\sin\frac{\pi}{3}\cos\frac{\pi}{6} + \cos\frac{\pi}{3}\sin\frac{\pi}{6}$ (b) $\tan\frac{\pi}{4} + \cot\frac{\pi}{4}$ (c) $\sec\frac{\pi}{3} \cdot \csc\frac{\pi}{6}$ .
Write down $\sin\frac{3\pi}{2}$ , $\cos\pi$ , $\tan\pi$ and $\cos\frac{\pi}{2}$ .
Is there an angle with $\sin\theta = 1.2$ ? Explain.
For $P (x, y)$ at distance $r$ from $O$ , show that $\sin^2\theta + \cos^2\theta = 1$ .
A Ferris wheel of radius 10 m has its centre 12 m above the ground. A seat that starts level with the centre, on the right, is at height $h = 12 + 10\sin\theta$ after turning through $\theta$ . Find $h$ for $\theta = \frac{\pi}{6}$ , $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

Show answersHide answers

$r = 13$ : $\sin\theta = -\frac{12}{13}$ , $\cos\theta = \frac{5}{13}$ , $\tan\theta = -\frac{12}{5}$ , $\csc\theta = -\frac{13}{12}$ , $\sec\theta = \frac{13}{5}$ , $\cot\theta = -\frac{5}{12}$ .
Quadrant III.
$\sin\theta = -\frac{4}{5}$ , $\tan\theta = \frac{4}{3}$ , $\csc\theta = -\frac{5}{4}$ .
(a) $\frac{3}{4} + \frac{1}{4} = 1$ (b) $1 + 1 = 2$ (c) $2 \times 2 = 4$ .
$- 1$ , $- 1$ , $0$ , $0$ .
No: $\sin\theta$ is the $y$ -coordinate of a point on a circle of radius 1, so $-1 \le \sin\theta \le 1$ .
$\sin^2\theta + \cos^2\theta = \frac{y^2}{r^2} + \frac{x^2}{r^2} = \frac{x^2 + y^2}{r^2} = \frac{r^2}{r^2} = 1$ .
17 m, 22 m and 2 m.

Quick check

In which quadrant is sin θ > 0 and cos θ < 0?

Key points

On the unit circle, $M(\cos\theta, \sin\theta)$ ; $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , and $\sec, \csc, \cot$ are the reciprocals of $\cos, \sin, \tan$ .
Signs by quadrant: A, S, T, C.
Exact values at $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$ come from the half equilateral triangle and the half square.

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the tables, the worked examples, the practice items and the questions.
Video: All 6 Trig Functions on the Unit Circle, Schattman Math, embedded from YouTube.

1Beyond the right-angled triangle

2Signs in the four quadrants

3Quadrantal and special angles

Beyond the right-angled triangle

Signs in the four quadrants

Quadrantal and special angles