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International Baccalaureate · DP 1: Applications and Interpretation HL

The Unit Circle and Radian Measure

⌚ ~130 min ▢ 14 blocks ✓ 62 exercises ➣ Prerequisites : Right-Angled Triangle Trigonometry, Unit Circle and Trigonometric Identities, Non-Right-Angled Trigonometry
I Radian Measure
I.1 Measuring Angles in Radians
The measure of an angle describes what fraction of a full revolution it represents. While degrees (360° in a circle) are a common unit, they are an arbitrary human invention. A more mathematically natural unit is the radian.
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Definition — Radian Measure
The radian measure of an angle \(\theta\) is the length of the arc it subtends on a circle of radius 1 centred at its vertex. When the direction of rotation matters, this measure is signed: positive for an anticlockwise rotation, negative for a clockwise one (see Positive and Negative Angles).
Figure
Proposition — Angle of a Full Circle
The circumference of a unit circle is \(C = 2\pi(1) = 2\pi\). Therefore, a full circle contains \(2\pi\) radians.
This establishes the fundamental conversion: \(360^\circ = 2\pi\) radians, which simplifies to \(180^\circ = \pi\) radians.
Figure
Method — Converting Between Degrees and Radians
Based on the relationship \(180^\circ = \pi\) radians:
  • To convert from degrees to radians, multiply by \(\dfrac{\pi}{180}\).
  • To convert from radians to degrees, multiply by \(\dfrac{180}{\pi}\).
Example
Convert \(60^{\circ}\) to radians.

\(\begin{aligned}[t] 60^{\circ}&=60^{\circ} \times \frac{\pi}{180^{\circ}}\\ &=\frac{\pi}{3} \end{aligned}\)

Definition — Positive and Negative Angles
  • A positive angle measure represents an anticlockwise rotation.
  • A negative angle measure represents a clockwise rotation.
Figure
Example
Draw the angles \(\dfrac{3\pi}{2}\) and \(-\dfrac{\pi}{2}\).

Figure

Method — Setting the Calculator Mode
A calculator computes cosines and sines in radians or in degrees, and nothing on the screen says which. Before any calculation, check the mode with a known value: \(\cos(1)\) is about \(0{.}54\) in radian mode and about \(0{.}9998\) in degree mode. When angles are measured in radians, the mode must be RAD.
Skills to practice
  • Converting Degrees to Radians in Terms of \(\pi\)
  • Converting Between Degrees and Radians
  • Calculating Sines and Cosines in Radian Mode
I.2 Arc Length and Sector Area
Proposition — Arc Length
On a circle of radius \(r\), an angle of \(\theta\) radians, with \(\theta > 0\), subtends an arc of length \(\ell = r\,\theta\). Equivalently, \(\theta = \dfrac{\ell}{r}\) and \(r = \dfrac{\ell}{\theta}\).
Figure

On the unit circle, the arc subtended by the angle \(\theta\) has length exactly \(\theta\): that is the definition of the radian.
Figure
The circle of radius \(r\) is the image of the unit circle under the enlargement (homothety) with centre \(O\) and scale factor \(r\): it multiplies every length by \(r\) and it carries the arc of the unit circle onto the arc that the same angle subtends. The two arcs are therefore proportional to the two radii:$$\frac{1}{r} = \frac{\theta}{\ell}, \qquad \text{hence} \qquad \ell = r\,\theta.$$ Check: a full turn, \(\theta = 2\pi\), gives \(\ell = 2\pi r\) — the whole circumference.

Example
A bicycle wheel has radius \(35\) cm and rolls without slipping. Through what distance does the bicycle move when the wheel turns by an angle of \(\dfrac{3\pi}{4}\)? Round to the nearest millimetre.

The wheel rolls without slipping: the distance travelled is the length of the arc swept.$$\begin{aligned}[t]\ell &= r\,\theta && \scriptscriptstyle\text{(arc length, angle in radians)} \\ &= 35 \times \frac{3\pi}{4} \\ &= \frac{105\pi}{4} \approx 82.5 \text{ cm.}\end{aligned}$$

Proposition — Area of a Sector in Radians
A sector is the part of a disc bounded by two radii and the arc between them. On a circle of radius \(r\), the sector of angle \(\theta\) radians, with \(0 < \theta \leq 2\pi\) — the sector whose arc is the one the angle \(\theta\) sweeps — has area \(A = \dfrac{1}{2}\,r^2\,\theta\). With its arc length \(\ell = r\,\theta\), this is also \(A = \dfrac{1}{2}\,\ell\,r\).
Figure

The area of a sector is proportional to its angle. Indeed, a rotation about the centre carries a sector onto any other sector with the same angle, so equal angles give equal areas; and putting two, three, … equal sectors side by side multiplies the angle and the area by the same number. A full turn measures \(2\pi\) radians (Angle of a Full Circle) and gives the whole disc, of area \(\pi r^2\). The sector of angle \(\theta\) is therefore the fraction \(\dfrac{\theta}{2\pi}\) of the disc: $$\begin{aligned}[t]A &= \frac{\theta}{2\pi} \times \pi r^2 && \text{(the fraction } \tfrac{\theta}{2\pi} \text{ of the disc)}\\ &= \frac{\theta\,r^2}{2} && \text{(simplifying by } \pi\text{)}\\ &= \frac{1}{2}\,r^2\,\theta.\end{aligned}$$ With the arc length \(\ell = r\,\theta\) (Arc Length), \(A = \dfrac{1}{2}\,r \times r\,\theta = \dfrac{1}{2}\,\ell\,r\). Check: a half-disc, \(\theta = \pi\), gives \(A = \dfrac{1}{2}\pi r^2\). In degrees the same fraction reads \(\dfrac{\theta}{360^\circ}\); in radians the full turn \(2\pi\) cancels the \(\pi\) of the disc, and the formula is shorter.

Example
A sector has radius \(6\) cm and angle \(\dfrac{2\pi}{3}\). Give the exact length of its arc and its exact area.

The angle is in radians, so both formulas apply directly. $$\begin{aligned}[t]\ell &= r\,\theta && \text{(arc length)}\\ &= 6 \times \frac{2\pi}{3} \\ &= 4\pi \text{ cm,}\end{aligned} \qquad \begin{aligned}[t]A &= \frac{1}{2}\,r^2\,\theta && \text{(area of a sector)}\\ &= \frac{1}{2} \times 36 \times \frac{2\pi}{3} \\ &= 12\pi \text{ cm}^2.\end{aligned}$$ Check with the arc: \(\dfrac{1}{2}\,\ell\,r = \dfrac{1}{2} \times 4\pi \times 6 = 12\pi\).

Example
A sector of radius \(5\) cm has area \(20 \text{ cm}^2\). Find its angle in radians, then its perimeter.

The area gives an equation in \(\theta\): $$\begin{aligned}[t]\frac{1}{2}\,r^2\,\theta &= A && \text{(area of a sector)}\\ \frac{1}{2} \times 25 \times \theta &= 20 && (r = 5)\\ 12.5\,\theta &= 20 \\ \theta &= \frac{20}{12.5} = 1.6 \text{ rad.}\end{aligned}$$ The boundary of the sector is made of two radii and the arc: $$\begin{aligned}[t]P &= 2r + r\,\theta && \text{(two radii and the arc)}\\ &= 10 + 5 \times 1.6 \\ &= 18 \text{ cm.}\end{aligned}$$ The angle \(1.6\) is not a multiple of \(\pi\): a measure in radians is a number like any other.

Skills to practice
  • Finding Exact Arc Lengths and Sector Areas
  • Finding an Angle or a Radius from a Sector
  • Justifying Arc and Sector Calculations
II The Unit Circle
II.1 Special Angles on the Circle
Recappreviously studied — click to reread
Up to now an angle was something one measures. From now on it will be a number. Imagine the number line as a thread, laid tangent to the circle of radius \(1\) at the point with \(x\)-coordinate \(1\), then wound around that circle: \(0\) stays where it is, the positive numbers wind anticlockwise, the negative numbers clockwise. Every real number then lands on a point of the circle.
Figure
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Definition — Image of a Real Number
The unit circle — the circle of radius \(1\) centred at the origin — is travelled anticlockwise: this is the positive direction. Winding the number line around it associates with every real number \(x\) a single point of the circle, called the image of \(x\) and written \(M(x)\): starting from the point of coordinates \((1, 0)\), one travels along the circle a distance \(|x|\), anticlockwise if \(x > 0\), clockwise if \(x < 0\). By the very definition of the radian, \(x\) is then a measure in radians of the signed angle between the positive half of the horizontal axis and the ray \(OM\), where \(M = M(x)\).
Proposition — Reals Having the Same Image
For every real number \(x\) and every integer \(k\):$$M(x + 2k\pi) = M(x).$$ Conversely, two real numbers have the same image on the unit circle if and only if their difference is a multiple of \(2\pi\).

The circle has radius \(1\), so one full turn has length \(2\pi\). Travelling an extra \(2\pi\) therefore brings the point back exactly where it was, and \(k\) extra turns bring it back \(k\) times: \(M(x + 2k\pi) = M(x)\). Conversely, suppose \(M(x) = M(y)\) with \(x \leq y\). Going from \(M(x)\) to \(M(y)\) one travels a distance \(y - x\) and comes back to the starting point: that distance is a whole number of turns, so \(y - x = 2k\pi\) for some integer \(k\).

The circle is cut into \(8\) equal arcs by \(4\) diameters. As the full circle measures \(2\pi\), each arc measures $$\frac{2\pi}{8}=\frac{\pi}{4}.$$
Figure
Counting the arcs anticlockwise from the angle \(0\), each arc adds \(\dfrac{\pi}{4}\): after \(k\) arcs, the angle is \(k \times \dfrac{\pi}{4}\).
Multiple of \(\dfrac{\pi}{4}\) Angle
\(1\times\dfrac{\pi}{4}=\dfrac{\pi}{4}\) Figure
\(2\times\dfrac{\pi}{4}=\dfrac{2\pi}{4}=\dfrac{\pi}{2}\) Figure
\(3\times\dfrac{\pi}{4}=\dfrac{3\pi}{4}\) Figure
\(4\times\dfrac{\pi}{4}=\dfrac{4\pi}{4}=\pi\) Figure
The count continues in the same way up to \(8\times\dfrac{\pi}{4}=\dfrac{8\pi}{4}=2\pi\), a full turn: the point is back at its starting position.
Four of these eight measures simplify to a fraction with a smaller denominator. The other four, \(\dfrac{\pi}{4}\), \(\dfrac{3\pi}{4}\), \(\dfrac{5\pi}{4}\) and \(\dfrac{7\pi}{4}\), are already in lowest terms.
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Proposition — Locating the Multiples of \(\frac{\pi}{4}\)
Figure
The circle is cut into \(12\) equal arcs by \(6\) diameters. As the full circle measures \(2\pi\), each arc measures $$\frac{2\pi}{12}=\frac{\pi}{6}.$$
Figure
Counting the arcs anticlockwise from the angle \(0\), each arc adds \(\dfrac{\pi}{6}\): after \(k\) arcs, the angle is \(k \times \dfrac{\pi}{6}\).
Multiple of \(\dfrac{\pi}{6}\) Angle
\(1\times\dfrac{\pi}{6}=\dfrac{\pi}{6}\) Figure
\(2\times\dfrac{\pi}{6}=\dfrac{2\pi}{6}=\dfrac{\pi}{3}\) Figure
\(3\times\dfrac{\pi}{6}=\dfrac{3\pi}{6}=\dfrac{\pi}{2}\) Figure
\(4\times\dfrac{\pi}{6}=\dfrac{4\pi}{6}=\dfrac{2\pi}{3}\) Figure
\(5\times\dfrac{\pi}{6}=\dfrac{5\pi}{6}\) Figure
\(6\times\dfrac{\pi}{6}=\dfrac{6\pi}{6}=\pi\) Figure
The count continues in the same way up to \(12\times\dfrac{\pi}{6}=\dfrac{12\pi}{6}=2\pi\), a full turn: the point is back at its starting position.
Eight of these twelve measures simplify to a fraction with a smaller denominator. The other four, \(\dfrac{\pi}{6}\), \(\dfrac{5\pi}{6}\), \(\dfrac{7\pi}{6}\) and \(\dfrac{11\pi}{6}\), are already in lowest terms.
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Proposition — Locating the Multiples of \(\frac{\pi}{6}\)
Figure
Proposition — Special Angles on the Unit Circle
Figure
Skills to practice
  • Identifying Angles on the Circle
II.2 Sine and Cosine on the Unit Circle
Recappreviously studied — click to reread
Every real number has a point on the circle. That point has two coordinates — and naming them is defining the cosine and the sine of a number, no longer of a triangle. For an acute angle, the right-angled triangle drawn under the point shows that nothing is lost: these are the very cosine and sine of the right-angled triangle.
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Definition — Unit circle
The unit circle is the circle with a radius of 1 centred at the origin: it is the circle used to measure angles, travelled anticlockwise.
Figure
Proposition — Relationship between Angle and Coordinates
For an angle \(\theta\), measured anticlockwise from the positive \(x\)-axis, \(M(x,y)\) is the corresponding point on the unit circle.
  • The \(x\)-coordinate is the cosine of the angle: \(\cos \theta = x\)
  • The \(y\)-coordinate is the sine of the angle: \(\sin \theta = y\)
For an acute angle, these equalities are proved with the right-angled triangle formed by \(O\), \(M\) and the foot of the perpendicular from \(M\) to the \(x\)-axis. For any other angle, these coordinates define \(\cos\theta\) and \(\sin\theta\).
Figure

Figure
For an acute angle \(\theta\), using the trigonometric ratios in the right-angled triangle \(OHM\): $$\begin{aligned} \cos \theta&=\frac{\text{adjacent}}{\text{hypotenuse}}\\ &=\frac{OH}{OM} \\ &=\frac{x}{1} \\ &=x \\ \end{aligned}$$ and $$\begin{aligned} \sin \theta&=\frac{\text{opposite}}{\text{hypotenuse}}\\ &=\frac{HM}{OM} \\ &=\frac{y}{1} \\ &=y \\ \end{aligned}$$

Example
Find the values \(\cos \left(\dfrac{\pi}{2}\right)\) and \(\sin \left(\dfrac{\pi}{2}\right)\).

On the unit circle, the point corresponding to the angle \(\dfrac{\pi}{2}\) has coordinates \((0,1)\):
Figure
\(\cos \left(\dfrac{\pi}{2}\right)=0 \quad\) \(x\)-coordinate
\(\sin \left(\dfrac{\pi}{2}\right)=1 \quad\) \(y\)-coordinate

Proposition — Sign of Sine and Cosine
Figure
Quadrant \(\cos \theta\) \(\sin \theta\)
I \(+\) \(+\)
II \(-\) \(+\)
III \(-\) \(-\)
IV \(+\) \(-\)
Skills to practice
  • Finding Sine and Cosine Values on the Unit Circle
III Identities and Exact Values
III.1 Trigonometric Identities
Recappreviously studied — click to reread
Once one point is placed on the circle, its symmetric points are placed too: a half-turn, a reflection in an axis, a reflection in the diagonal each send \(M(\theta)\) onto another point of the circle. Reading each of these symmetries on the coordinates costs nothing — and turns a handful of exact values into a whole table: that is what the identities below do.
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Proposition — Pythagorean Identity
For any angle \(\theta\): $$ \cos^2 \theta + \sin^2 \theta = 1 $$

Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\), and \(H\) the foot of the perpendicular from \(M\) to the \(x\)-axis.
Figure
The \(x\)-coordinate of \(H\) is \(\cos\theta\), and the length \(OH\) is its absolute value: \(OH = |\cos\theta|\); in the same way, \(HM = |\sin\theta|\). By Pythagoras' theorem in the triangle \(OHM\), right-angled at \(H\): $$\begin{aligned}[t] OH^2+HM^2&=OM^2 && \text{(Pythagoras' theorem)}\\ |\cos \theta|^2+|\sin \theta|^2&=1^2 && (OM = 1)\\ \cos ^2 \theta+\sin ^2 \theta&=1 && \text{(a square does not depend on the sign)} \end{aligned}$$ When \(M\) lies on an axis, the triangle is flat: one coordinate is \(0\), the other is \(1\) or \(-1\), and the equality still holds. On the figure \(\theta\) is acute, and the two coordinates are the lengths \(OH\) and \(HM\) themselves.

Proposition — Periodicity Identity
For any angle \(\theta\) and any integer \(k\): $$ \cos(\theta + 2k\pi) = \cos\theta \quad \text{and} \quad \sin(\theta + 2k\pi) = \sin\theta $$

Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\).
Let \(M'(\cos(\theta+2\pi),\sin(\theta+2\pi))\) be the point on the unit circle at the angle \(\theta+2\pi\).
Figure
By the proposition on the reals having the same image, \(M(\theta+2\pi)=M(\theta)\): the two points coincide on the circle, \(M'=M\).
Thus, \(\cos (\theta+2 \pi)=\cos \theta\) and \(\sin (\theta+2 \pi)=\sin \theta\).
The same argument holds for any multiple of \(2\pi\), i.e. for any integer \(k\).

Proposition — Half-Turn: \(\pi + \theta\)
Reflection through the origin: $$\begin{aligned} \sin (\pi+\theta)&=-\sin \theta \\ \cos (\pi+\theta)&=-\cos \theta \end{aligned}$$
Figure

Let \(\theta\) be an angle.
Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\).
Let \(M'(\cos(\pi+\theta),\sin(\pi+\theta))\) be the point on the unit circle at the angle \(\pi+\theta\).
A rotation by an angle \(\pi\) is a half-turn about the origin \(O\), which is the same as a point reflection through \(O\). Therefore, the coordinates of \(M'\) are the opposites of the coordinates of \(M\).
So \(\begin{aligned}[t] \sin (\pi+\theta)&=-\sin \theta \\ \cos (\pi+\theta)&=-\cos \theta \end{aligned}\)

Proposition — Symmetry about the Horizontal Axis: \(-\theta\)
Reflection in the \(x\)-axis: $$\begin{aligned} \sin (-\theta)&=-\sin \theta \\ \cos (-\theta)&=\cos \theta \end{aligned}$$
Figure

Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\).
Let \(M'(\cos(-\theta),\sin(-\theta))\) be the point on the unit circle at the angle \(-\theta\).
The point \(M'\) is the reflection of \(M\) in the \(x\)-axis. Therefore, the \(x\)-coordinate of \(M'\) is the same as that of \(M\), and the \(y\)-coordinate of \(M'\) is the opposite of the \(y\)-coordinate of \(M\).
So \(\begin{aligned}[t] \sin (-\theta)&=-\sin \theta \\ \cos (-\theta)&=\cos \theta \end{aligned}\)

Proposition — Symmetry about the Vertical Axis
For every real number \(\theta\):$$\cos(\pi - \theta) = -\cos\theta, \qquad \sin(\pi - \theta) = \sin\theta.$$ The points of angles \(\theta\) and \(\pi - \theta\) are mirror images of each other in the vertical axis: they share the same \(y\)-coordinate, and their \(x\)-coordinates are opposite.
Figure

Write \(\pi - \theta = \pi + (-\theta)\) and apply the two preceding propositions, first the one for the opposite, then the one for adding \(\pi\):$$\begin{aligned}[t] \cos(\pi - \theta) &= \cos\big(\pi + (-\theta)\big) \\ &= -\cos(-\theta) && \scriptscriptstyle\text{(adding } \pi \scriptscriptstyle\text{)} \\ &= -\cos\theta && \scriptscriptstyle\text{(cosine of the opposite)} \end{aligned}$$and, in the same way,$$\begin{aligned}[t] \sin(\pi - \theta) &= \sin\big(\pi + (-\theta)\big) \\ &= -\sin(-\theta) && \scriptscriptstyle\text{(adding } \pi \scriptscriptstyle\text{)} \\ &= -(-\sin\theta) = \sin\theta. && \scriptscriptstyle\text{(sine of the opposite)} \end{aligned}$$

Proposition — Identities with \(\frac{\pi}{2} - \theta\)
Reflection in the line \(y=x\): $$\begin{aligned} \cos\left(\frac \pi 2 -\theta\right) &= \sin \theta \\ \sin\left(\frac \pi 2 -\theta\right) &= \cos \theta \\ \end{aligned}$$
Figure

Let \(\theta\) be an angle.
Let \(M(\cos \theta,\sin\theta)\) be the point on the unit circle at the angle \(\theta\).
Let \(M'(\cos(\frac \pi 2 -\theta),\sin(\frac \pi 2 -\theta))\) be the point on the unit circle at the angle \(\frac \pi 2 -\theta\).
The points \(M\) and \(M'\) are symmetric with respect to the line \(y=x\), so the \(x\)-coordinate of \(M\) is the \(y\)-coordinate of \(M'\), and the \(y\)-coordinate of \(M\) is the \(x\)-coordinate of \(M'\).
So \(\begin{aligned}[t] \cos\left(\frac \pi 2 -\theta\right) &= \sin \theta \\ \sin\left(\frac \pi 2 -\theta\right) &= \cos \theta \\ \end{aligned}\)

Example
Using the identities above, write:
  1. \(\cos\dfrac{2\pi}{3}\) in terms of \(\cos\dfrac{\pi}{3}\);
  2. \(\sin\dfrac{7\pi}{6}\) in terms of \(\sin\dfrac{\pi}{6}\).

  1. \(\dfrac{2\pi}{3} = \pi - \dfrac{\pi}{3}\): the symmetry about the vertical axis changes the sign of the \(x\)-coordinate, so$$\cos\frac{2\pi}{3} = -\cos\frac{\pi}{3}.$$
  2. \(\dfrac{7\pi}{6} = \pi + \dfrac{\pi}{6}\): the half-turn about \(O\) changes the sign of both coordinates, so$$\sin\frac{7\pi}{6} = -\sin\frac{\pi}{6}.$$
In each case the first step is the same: decompose the angle so as to read a symmetry of the circle on it.

Skills to practice
  • Finding Exact Values with the Pythagorean Identity
III.2 Multiples of \(\frac{\pi}{4}\)
Recappreviously studied — click to reread
The symmetries of the circle relate the coordinates of many points to those of a single one — but they give nothing until a first value is known. Here is the first: the eighth of a turn, \(\dfrac{\pi}{4}\), that is \(45^\circ\), where the point sits on the diagonal. Its two coordinates are equal, and the right-angled triangle drawn under the point is enough to find them: it is the isosceles right-angled triangle that already gives the exact values of the ratios of \(45^\circ\).
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Proposition — Coordinates for Angle \(\frac{\pi}{4}\)
$$\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2} \quad \text { and } \quad \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$$
Figure

Figure
The sum of the angles in a triangle is \(\pi\), so \(\angle{OMH}=\pi - \dfrac \pi 4 - \dfrac \pi 2 = \dfrac \pi 4\).
Since \(\angle{OMH}=\angle{MOH}\), the triangle \(OHM\) is isosceles.
Let \(a=OH=HM\). $$\begin{aligned} a^2+a^2 &=1^2 \quad \text { (Pythagoras' theorem in the right-angled triangle } OHM)\\ 2 a^2 &=1 \\ a^2 &=\frac{1}{2} \\ a &=\frac{\sqrt{2}}{2} \quad \text { since } a\geq 0 \end{aligned}$$ So \(M\left(\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)\).
As \(\cos \theta\) is the \(x\)-coordinate of \(M\) and \(\sin \theta\) is the \(y\)-coordinate of \(M\), $$\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2} \quad \text { and } \quad \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$$

Proposition — Multiples of \(\frac{\pi}{4}\)
Figure

The coordinates of each point are found by using the symmetries about the axes and the half-turn about \(O\).

Example
Find \(\cos \dfrac{3 \pi}{4}\).

\(\cos \dfrac{3 \pi}{4} = -\dfrac{\sqrt{2}}{2}\)

Skills to practice
  • Finding an Exact Value with an Identity
III.3 Multiples of \(\frac{\pi}{6}\)
Recappreviously studied — click to reread
A second family: the twelfth of a turn, \(\dfrac{\pi}{6}\) (\(30^\circ\)), and its companion \(\dfrac{\pi}{3}\) (\(60^\circ\)), the sixth of a turn. This time the triangle to draw is equilateral: its half is the right-angled triangle that already gives the exact values of the ratios of \(30^\circ\) and \(60^\circ\). It gives the two angles at once — they exchange their coordinates.
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Proposition — Coordinates of the Angles \(\frac{\pi}{3}\) and \(\frac{\pi}{6}\)
$$\cos \frac{\pi}{3}=\frac{1}{2} \quad \text { and } \quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}, \qquad\qquad \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \quad \text { and } \quad \sin \frac{\pi}{6}=\frac{1}{2}.$$ The two angles exchange their coordinates: the point at \(\frac{\pi}{6}\) is the mirror image of the point at \(\frac{\pi}{3}\) in the line \(y = x\).
Figure

Figure
Let \(\angle{MON}=\dfrac \pi 3\).
As \(ON=OM=1\), the triangle \(OMN\) is isosceles at \(O\): its base angles are equal, \(\angle{OMN}=\angle{ONM}\).
Since the sum of the angles in a triangle is \(\pi\), \(2\,\angle{OMN}=\pi-\dfrac \pi 3=\dfrac{2\pi}{3}\), so \(\angle{OMN}=\angle{ONM}=\dfrac \pi 3\).
So the triangle \(OMN\) is equilateral.
The altitude \(MH\) bisects the base \(ON\).
Thus \(OH=\dfrac{1}{2}\). $$\begin{aligned} OH^2+HM^2 &=OM^2 \quad \text { (Pythagoras' theorem in the right-angled triangle } OHM)\\ \left(\frac{1}{2}\right)^2 + HM^2 &=1 \\ HM^2 &=\frac{3}{4} \\ HM &=\frac{\sqrt{3}}{2} \quad \text { since } HM\geq 0 \end{aligned}$$ As \(\cos \theta\) is the \(x\)-coordinate of \(M\) and \(\sin \theta\) is the \(y\)-coordinate of \(M\): $$\cos \frac{\pi}{3}=\frac{1}{2} \quad \text { and } \quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$$ By the symmetry about the line \(y=x\) (the identities with \(\frac{\pi}{2}-\theta\)), the twelfth of a turn, \(\frac{\pi}{6}\), follows at once:$$\cos \frac{\pi}{6}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \quad \text { and } \quad \sin \frac{\pi}{6}=\cos \frac{\pi}{3}=\frac{1}{2}.$$

Proposition — Multiples of \(\frac{\pi}{6}\)
Figure

The coordinates of each point are found by applying the symmetries about the axes and the half-turn about \(O\).

Example
Find \(\cos \dfrac{2 \pi}{3}\) and \(\sin \dfrac{2 \pi}{3}\).

Figure
\(\cos \dfrac{2 \pi}{3}=-\dfrac{1}{2}\) and \(\sin \dfrac{2 \pi}{3}=\dfrac{\sqrt{3}}{2}\)

Proposition — Table of Exact Values
\(\theta\) \(0\) \(\dfrac{\pi}{6}\) \(\dfrac{\pi}{4}\) \(\dfrac{\pi}{3}\) \(\dfrac{\pi}{2}\)
\(\cos\theta\) \(1\) \(\dfrac{\sqrt3}{2}\) \(\dfrac{\sqrt2}{2}\) \(\dfrac{1}{2}\) \(0\)
\(\sin\theta\) \(0\) \(\dfrac{1}{2}\) \(\dfrac{\sqrt2}{2}\) \(\dfrac{\sqrt3}{2}\) \(1\)
Every multiple of \(\dfrac{\pi}{6}\) and of \(\dfrac{\pi}{4}\) is brought back to this table by the symmetries of the circle: a whole turn subtracted, the opposite, the half-turn, the symmetry about the vertical axis. Read from left to right, the cosines decrease from \(1\) to \(0\) while the sines increase from \(0\) to \(1\) — the two rows are mirror images of each other.
Skills to practice
  • Finding Exact Values with Symmetries
IV The Tangent
IV.1 The Tangent on the Unit Circle
Definition — Tangent Function
The tangent of an angle \(\theta\) is defined, whenever \(\cos \theta \neq 0\), as the ratio of the sine to the cosine: $$\tan \theta = \frac{\sin \theta}{\cos \theta}.$$ Equivalently, \(\tan \theta\) is defined for all real \(\theta\) such that \(\theta \neq \dfrac{\pi}{2} + k\pi\) for any integer \(k\).
Proposition — Geometric Interpretation of Tangent
On the unit circle, for any angle \(\theta\) with \(\cos \theta \neq 0\), the line through the origin forming an angle \(\theta\) with the positive \(x\)-axis meets the vertical tangent line \(x = 1\) at the point \((1,\tan \theta)\). In particular, \(\tan \theta\) is the \(y\)-coordinate of this intersection point.
Figure

The ray from the origin at an angle \(\theta\) passes through the point \(P(\cos \theta, \sin \theta)\) on the unit circle. Provided \(\cos \theta \neq 0\), the gradient of the line through the origin and \(P\) is $$\frac{\sin \theta}{\cos \theta} = \tan \theta,$$ so the equation of this line is \(y = (\tan \theta)\,x\). The intersection of this line with the vertical line \(x = 1\) occurs when \(y = (\tan \theta)\cdot 1 = \tan \theta\). Thus, the intersection point is \((1,\tan \theta)\).

Proposition — Tangent Values for Common Angles
\(\theta\) \(\sin \theta\) \(\cos \theta\) \(\tan \theta\)
\(0\) \(0\) \(1\) \(0\)
\(\dfrac{\pi}{6}\) \(\dfrac{1}{2}\) \(\dfrac{\sqrt{3}}{2}\) \(\dfrac{\sqrt{3}}{3}\)
\(\dfrac{\pi}{4}\) \(\dfrac{\sqrt{2}}{2}\) \(\dfrac{\sqrt{2}}{2}\) \(1\)
\(\dfrac{\pi}{3}\) \(\dfrac{\sqrt{3}}{2}\) \(\dfrac{1}{2}\) \(\sqrt{3}\)
\(\dfrac{\pi}{2}\) \(1\) \(0\) undefined
The values in other quadrants follow from the symmetries of the unit circle and the fact that sine and cosine are \(2\pi\)-periodic.
Skills to practice
  • Evaluating the Tangent at Special Angles
V Going further
Skills to practice
  • Solving Problems with Arcs and Sectors
  • Reasoning on the Unit Circle
  • Quiz