3.5 The unit circle

In right-angled triangles, sine, cosine and tangent are ratios of sides, so they only make sense for angles between $0$ and $\frac{\pi}{2}$ ($0°$ and $90°$). Many situations involve larger angles: the obtuse angle of a triangle, the position of a point on a turning wheel after several revolutions, or a direction measured clockwise. The unit circle extends the three ratios to every angle, positive or negative, in a way that agrees with the right-angled triangle definitions for acute angles.

This extended definition underlies the graphs of the circular functions, the solution of trigonometric equations and the identities that link the ratios. It also explains why the sine rule can give two different triangles from the same information.

Angles of any size

Measuring angles from the positive x-axis

On a coordinate grid, an angle is measured from the positive $x$-axis. A radius starts along the positive $x$-axis and turns about the origin. An anticlockwise turn gives a positive angle, and a clockwise turn gives a negative angle. A turn of more than $2\pi$ (one full revolution) is also allowed: an angle of $\frac{5\pi}{2}$ is one full turn followed by a further $\frac{\pi}{2}$.

The axes divide the plane into four quadrants, numbered anticlockwise starting from the top right.

Angles that differ by a whole number of turns end in the same position. For example, $\frac{\pi}{3}$, $\frac{7\pi}{3}$ and $-\frac{5\pi}{3}$ all leave the radius pointing in the same direction, because $\frac{7\pi}{3} = \frac{\pi}{3} + 2\pi$ and $-\frac{5\pi}{3} = \frac{\pi}{3} - 2\pi$.

The unit circle definitions

Sine and cosine as coordinates

The unit circle is the circle of radius $1$ with its centre at the origin. When a radius turns through an angle $\theta$, its end point $P$ lies on this circle.

For an acute angle, this definition agrees with the ratios from right-angled triangles. Dropping a perpendicular from $P$ to the $x$-axis gives a right-angled triangle with hypotenuse $1$. Its adjacent side is $\frac{\text{adjacent}}{1} = \cos\theta$ and its opposite side is $\frac{\text{opposite}}{1} = \sin\theta$. The new definition gives the same values for acute angles and also works for every other angle, because every angle has an end point on the circle.

Because every point of the unit circle has coordinates between $-1$ and $1$,

for every angle $\theta$. An equation such as $\sin\theta = 1.2$ has no solution.

Tangent

The tangent of an angle is defined from the sine and cosine.

Since $P$ is $(\cos\theta, \sin\theta)$ and $O$ is the origin, $\frac{\sin\theta}{\cos\theta}$ is the rise over the run from $O$ to $P$. So $\tan\theta$ is the gradient of the radius $OP$. This gives a quick picture of its behaviour: the gradient is positive when $OP$ slopes upwards to the right (first and third quadrants), negative when it slopes downwards (second and fourth quadrants), and undefined when $OP$ is vertical, at $\theta = \frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$. Unlike sine and cosine, $\tan\theta$ can take any real value.

The ratios at the axes

When $P$ lies on an axis, its coordinates can be read straight from the circle.

Points on a circle of any radius

A circle of radius $r$ centred at the origin is the unit circle enlarged by a scale factor $r$. The point at angle $\theta$ on it is therefore

This describes the position of anything moving in a circle, such as a capsule on an observation wheel or the tip of a rotating blade, and it is the starting point for the circular function models in 3.7.

Exact values from special triangles

Two special triangles

The trigonometric ratios of most angles are irrational numbers that a calculator gives only approximately. For the angles $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{\pi}{3}$ ($30°$, $45°$ and $60°$) the ratios have exact values involving simple surds, which come from two triangles.

The first triangle is half of an equilateral triangle with sides $2$. An equilateral triangle has three angles of $\frac{\pi}{3}$. Cutting it in half along a line of symmetry gives a right-angled triangle with hypotenuse $2$, shortest side $1$ and angles $\frac{\pi}{6}$ and $\frac{\pi}{3}$. Pythagoras's theorem gives the third side: $\sqrt{2^2 - 1^2} = \sqrt{3}$.

The second triangle is an isosceles right-angled triangle with two sides of $1$. Its other two angles are equal, so each is $\frac{\pi}{4}$, and its hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$.

Reading the ratios from these triangles gives the table of exact values. The values at $0$ and $\frac{\pi}{2}$ come from the unit circle.

The value $\frac{\sqrt{2}}{2}$ is the same as $\frac{1}{\sqrt{2}}$, obtained by rationalising the denominator; both forms are correct.

The patterns in the table have a reason. Since $\frac{\pi}{6}$ and $\frac{\pi}{3}$ are the two acute angles of the same right-angled triangle, the side opposite one is adjacent to the other, so $\sin\frac{\pi}{6} = \cos\frac{\pi}{3}$ and $\cos\frac{\pi}{6} = \sin\frac{\pi}{3}$. In general, $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$ for complementary angles; the "co" in cosine stands for "complement".

Signs of the ratios in each quadrant

The CAST diagram

The sign of each ratio depends only on the quadrant in which $P$ lies. Since $\cos\theta$ is the $x$-coordinate, it is positive to the right of the $y$-axis (first and fourth quadrants). Since $\sin\theta$ is the $y$-coordinate, it is positive above the $x$-axis (first and second quadrants). And $\tan\theta = \frac{\sin\theta}{\cos\theta}$ is positive when the two have the same sign (first and third quadrants).

Related angles

Symmetry of the unit circle

The unit circle is symmetric in both axes and about the origin. A point $P$ at angle $\theta$ in the first quadrant therefore has three mirror images, one in each other quadrant, each making the same acute angle $\theta$ with the $x$-axis. This acute angle is the related angle (also called the reference angle).

Reading the coordinates from the diagram gives the related angle results. Each row follows from the position of the point and the CAST signs.

These results hold for every value of $\theta$, not only acute ones; the quadrant column describes the picture when $\theta$ is acute. The angle $-\theta$ ends at the same point as $2\pi - \theta$, so its row matches. In degrees the same results read $\sin(180° - \theta) = \sin\theta$, $\cos(180° - \theta) = -\cos\theta$, and so on.

Finding the ratio of any angle

To find an exact value for any angle, work through three steps:

1. Find the quadrant in which the angle ends (subtract multiples of $2\pi$ first if the angle is larger than $2\pi$ or negative).

2. Find the related angle, the acute angle to the $x$-axis.

3. Take the ratio of the related angle and attach the sign given by CAST.

Angles outside the interval $0$ to $2\pi$ are handled in the same way after adding or subtracting whole turns, since a whole turn returns $P$ to the same point.

Expressions with an unknown angle

The related angle results also simplify expressions in which the angle is a letter.

Two angles with the same ratio

Reversing the process, each value of a ratio between $-1$ and $1$ (other than $\pm 1$) comes from exactly two angles in $0 \le \theta < 2\pi$: one in each of the two quadrants where the ratio has that sign. The graph of $y = \sin x$ shows this for $\sin x = \frac{1}{2}$: the line $y = \frac{1}{2}$ meets the curve at $x = \frac{\pi}{6}$ and $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

The ambiguous case of the sine rule

Why two triangles can fit

The sine rule from 3.2 finds an unknown angle from its sine:

Because $\sin(180° - \theta) = \sin\theta$, a value such as $\sin B = 0.8333$ is produced by an acute angle $B \approx 56.4°$ and also by the obtuse angle $180° - 56.4° = 123.6°$. A calculator gives only the acute one. Sometimes both angles fit a triangle with the given measurements; this is the ambiguous case.

The diagram shows the geometry. The side $AC$ and the angle at $A$ are fixed. The third side, $CB = 6$ cm, can swing like a hinge about $C$, and its end meets the base line at two points. The two triangles $AB_1C$ and $AB_2C$ both have $A = 30°$, $AC = 10$ cm and $BC = 6$ cm. The triangle $B_1CB_2$ is isosceles, so the angles at $B_1$ and $B_2$ are supplementary: one is acute and the other obtuse.

When does the ambiguous case occur?

Suppose angle $A$ (acute), side $b = AC$ and side $a = BC$, opposite $A$, are given. The distance from $C$ to the base line is $b\sin A$. Comparing $a$ with this distance and with $b$ decides how many triangles exist.

In practice, there is no need to memorise this table. Find the acute angle from the sine rule, form the obtuse alternative $180° - \theta$, and check whether the angles of each triangle add up to less than $180°$.

The obtuse alternative does not always survive the check, as the next example shows.

The conditions can also be used in reverse, to find which lengths give two triangles.

Exam-style questions on the unit circle

Questions on this content come in two kinds: short non-calculator questions on exact values and signs, and longer calculator questions in which the sine rule gives two triangles.

Self-check

• To do:

  I can define $\sin\theta$ and $\cos\theta$ as the coordinates of a point on the unit circle, and $\tan\theta$ as $\frac{\sin\theta}{\cos\theta}$, the gradient of the radius.

• To do:

  I can derive the exact values for $\frac{\pi}{6}$, $\frac{\pi}{4}$ and $\frac{\pi}{3}$

• To do:

  I can state the sign of each ratio in each quadrant using CAST.

• To do:

  I can find the exact value of a ratio of any angle, including negative angles and angles larger than $2\pi$, using related angles.

• To do:

  I can simplify expressions such as $\sin(\pi - \alpha)$, $\cos(\pi + \alpha)$ and $\tan(2\pi - \alpha)$

• To do:

  I can find the coordinates $(r\cos\theta, r\sin\theta)$ of a point on a circle of radius $r$.

• To do:

  I can decide whether the sine rule gives zero, one or two triangles, and find all the possible triangles.

Find the exact length of $[\mathrm{DC}]$.

[4]4 marks

The angle is in the second quadrant, so $Q$ should be to the left of the $y$-axis and above the $x$-axis, which it is.

Since $\sqrt{3} \times \sqrt{3} = 3$:

(b) $\frac{4\pi}{3} = \pi + \frac{\pi}{3}$, in the third quadrant, where cosine is negative. The related angle is $\frac{\pi}{3}$.

(c) $\frac{7\pi}{4} = 2\pi - \frac{\pi}{4}$, in the fourth quadrant, where tangent is negative. The related angle is $\frac{\pi}{4}$.

(d) $135° = 180° - 45°$, in the second quadrant, where cosine is negative.

(b) Subtract one full turn: $\frac{13\pi}{6} - 2\pi = \frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6}$.

(c) Subtract one full turn: $\frac{11\pi}{3} - 2\pi = \frac{5\pi}{3} = 2\pi - \frac{\pi}{3}$, in the fourth quadrant, where tangent is negative.

(c) $\tan(2\pi - \alpha) = -\tan\alpha = -\frac{0.6}{0.8} = -0.75$

(d) $\sin(-\alpha) + \cos(-\alpha) = -\sin\alpha + \cos\alpha = -0.6 + 0.8 = 0.2$

Third quadrant: $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

So $\theta = \frac{2\pi}{3}$ or $\theta = \frac{4\pi}{3}$. Equations of this kind, over other intervals and with more complicated angles, are the subject of 3.8.

Check both: $30° + 123.6° = 153.6° < 180°$, so both triangles exist.

Case 1, $B = 56.44°$: $C = 180° - 30° - 56.44° = 93.56°$.

Case 2, $B = 123.56°$: $C = 180° - 30° - 123.56° = 26.44°$.

These match the diagram: $AB_2 \approx 12.0$ cm with an acute angle at $B_2$, and $AB_1 \approx 5.34$ cm with an obtuse angle at $B_1$.

Test the obtuse value: $40° + 147.6° = 187.6° > 180°$, which is impossible in a triangle. So only one triangle exists, and

This agrees with the table: here $a = 12 \ge b = 10$, so there is exactly one triangle.

For $x = 5$ there is one right-angled triangle, for $x < 5$ there is none, and for $x \ge 10$ there is one.

(b) "Hence" means use part (a).

(c) $\frac{5\pi}{6} = \pi - \frac{\pi}{6}$ (second quadrant, tangent negative) and $\frac{4\pi}{3} = \pi + \frac{\pi}{3}$ (third quadrant, sine negative).

So $k = -\frac{5}{6}$.

Check the obtuse value: $45° + 114.61° = 159.61° < 180°$, so both triangles exist. The two values are $65.4°$ and $114.6°$.

(b) With $R = 114.613\ldots°$, the third angle is $Q = 180° - 45° - 114.613\ldots° = 20.386\ldots°$.

(c) For the acute case, $R = 65.386\ldots°$ and $Q = 69.613\ldots°$, so

Using area $= \frac{1}{2} \times PQ \times PR \times \sin 45°$ for each triangle:

[3]3 marks