Trigonometric Functions - Maths: an open, verified university mathematics reference

Why this matters¶

A Ferris wheel of radius 20 metres turns at a steady speed, once every 10 minutes, and its centre is 22 metres above the ground. You board at the bottom. How high are you after 100 seconds? After 25 minutes?

Your height goes up and down between 2 and 42 metres and repeats every 10 minutes. To describe it we need two functions that say where a point is after it has turned some way round a circle: how far it is across, and how high. They are the cosine and the sine. The same two functions describe everything that turns or swings back and forth: wheels and pendulums, tides, sound waves and alternating current. We answer the question about the wheel in Example 6.

Calculus adds one requirement: an angle is measured by a length, the length of an arc of a circle of radius 1. This unit, the radian, is what makes the formulas of calculus for sine and cosine simple, and this page builds everything on it.

Angles in radians¶

Everything on this page rests on a few facts of plane geometry that you know from school: they are the starting point, not something this course proves. They are used directly in these places, and through them by every example and exercise:

the definitions of the radian (Definition 1) and of sine and cosine (Definition 2), and the arc $r\theta$ and the sector $\frac12 r^2\theta$ on a circle of radius $r$ after them;
the reasons for the properties in Remark 2, and the “In words” of Definition 3 (the line through $O$ and $P(t)$ );
the proofs of Theorem 1, Theorem 2, Lemma 1 and Proposition 4, and the figure of the lemma;
the Values bullet of Graphs: period, amplitude and phase;
the worked examples on a wheel (Example 1) and on the Ferris wheel (Example 6);
the exercises on a pendulum (Exercise 2), on solving $\sin t = -\frac12$ (Exercise 7, its hint and solution) and on the Ferris wheel (Exercise 9, its solution).

They use only the facts in the box below.

Remark 1 (Facts from school used on this page)

We write $O = (0, 0)$ for the origin and $A = (1, 0)$ .

Distance. By Pythagoras’ theorem, the distance between the points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ . So the unit circle, the circle of radius 1 about $O$ , is the set of points $(x, y)$ with $x^2 + y^2 = 1$ , and the unit disc is the set of points with $x^2 + y^2 \le 1$ . The disc contains, with any two of its points, the segment between them.
Length of arcs. Every arc of the unit circle has a length. An arc cut into two arcs at a point has the sum of their lengths. The whole circle has length $2\pi$ : this is the definition of the number $\pi$ .
Journeys along the circle. From any point of the circle, travelling a distance $d \ge 0$ along it anticlockwise, or clockwise, ends at exactly one point; anticlockwise is the direction in which a journey from $A$ sets off towards $(0, 1)$ . Counting clockwise distances as negative, a journey of $s$ followed by a journey of $t$ ends where a journey of $s + t$ does. Every point of the circle is reached from $A$ by an anticlockwise journey of exactly one length in $[0, 2\pi)$ .
Journeys and arcs. An anticlockwise journey of length $d$ , where $0 < d \le 2\pi$ , runs along an arc of the circle from its starting point to its end point, and that arc has length $d$ . (For $d = 2\pi$ the arc is the whole circle.)
Rotations and reflections. For every point $Q$ of the unit circle there is a rotation about $O$ that maps $A$ to $Q$ . A rotation about $O$ , and a reflection in a line through $O$ , map the unit circle onto itself and keep the distances between points and the lengths of arcs. So they map a journey along the circle to a journey of the same length; a rotation keeps its direction, and a reflection reverses it.
Circles of other radii. A circle of radius $r > 0$ is the unit circle scaled by $r$ (and moved): lengths of arcs are multiplied by $r$ , and areas by $r^2$ .
Sectors. Take an arc of the unit circle of length $\theta$ , where $0 < \theta \le 2\pi$ . The region between the radii at its ends is the set of points $\lambda X$ with $\lambda \ge 0$ and $X$ on the arc, and the sector of the arc is the part of this region that lies in the unit disc. If $\theta < \pi$ , the region contains, with any two of its points, the segment between them; and if $Q$ and $R$ are points other than $O$ on the two rays from $O$ through the ends of the arc, the triangle $OQR$ is the part of the region that lies on the same side of the line $QR$ as $O$ , or on that line. Every point of that triangle lies on a segment from $O$ to a point of its side $QR$ .
Area. A triangle with base $b$ and height $h$ has area $\frac12 bh$ . A sector whose arc has length $\theta$ , where $0 < \theta \le 2\pi$ , has area $\frac{\theta}{2}$ . For $\theta = 2\pi$ this says that the unit disc has area $\pi$ : that is Archimedes’ theorem, not the definition of $\pi$ , which is the length of the circle. A region that lies inside another has at most the other’s area.

No page of this course proves these facts. They are proved in Real Analysis (ana), outside this course.

To measure an angle at the centre of the unit circle, walk along the circle, inside the angle, from one arm to the other and measure how far you went. The larger the angle, the longer the walk.

In words. An angle of $\theta$ radians is the angle you turn through while you walk a distance $\theta$ along the unit circle. Radians are lengths divided by the radius, so the unit is usually left out: “an angle of 2” means 2 radians.

Example. A full turn is the whole circle, $2\pi$ radians. A straight angle is half of it, $\pi$ , and a right angle is a quarter, $\frac{\pi}{2}$ (property 1 of Remark 2 shows that the quarter circles have equal lengths). A full turn is also 360 degrees, so 1 degree is $\frac{2\pi}{360} = \frac{\pi}{180}$ radians: 60 degrees is $\frac{60\pi}{180} = \frac{\pi}{3}$ radians, and an angle of $\theta$ radians is $\frac{180\theta}{\pi}$ degrees.

Non-example. An angle of 1 radian is not 1 degree. It is the angle whose arc is exactly as long as the radius, $\frac{180}{\pi} \approx 57.3$ degrees.

On a circle of radius $r$ , an angle of $\theta$ radians at the centre, with $0 < \theta \le 2\pi$ , cuts off an arc of length $r\theta$ and a sector of area $\frac12 r^2 \theta$ , by the facts on circles of other radii and on areas. For $\theta = 2\pi$ these are the circumference $2\pi r$ and the area $\pi r^2$ of the whole disc.

Sine and cosine¶

Now let the angle grow beyond a full turn, and let it turn the other way too. Start at $A = (1, 0)$ and walk along the unit circle; after a distance $t$ you are at a point $P(t)$ . Its coordinates are the cosine and the sine of $t$ .

The unit circle about the origin O, with the x- and y-axes. The x-axis is marked −1 on the left, where the circle crosses it, and A on the right; the y-axis is marked 1 and −1. A thick arc runs anticlockwise along the circle from the point A = (1, 0) to a point P(t) = (cos t, sin t) in the first quadrant and is labelled "arc length t". A radius joins O to P(t). A thick segment on the x-axis from O to the foot of P(t) is labelled cos t, and a dashed vertical segment from there up to P(t) is labelled sin t. — Figure 1:The point $P(t)$ is reached from $A = (1, 0)$ by walking a distance $t$ along the unit circle. Its $x$ -coordinate is $\cos t$ and its $y$ -coordinate is $\sin t$ .

In words. $\cos t$ is how far across, and $\sin t$ how high, you are after walking a distance $t$ round the unit circle from $A$ . For $0 < t < \frac{\pi}{2}$ they are the two legs of the right-angled triangle with hypotenuse $OP(t)$ in Figure 1, which is the definition from school (“adjacent over hypotenuse” and “opposite over hypotenuse”, with hypotenuse 1). The unit circle extends it to every real number $t$ , and $t$ is always in radians.

Example. $P(0) = A = (1, 0)$ , so $\cos 0 = 1$ and $\sin 0 = 0$ . After a quarter of the circle, $P\bigl(\frac{\pi}{2}\bigr) = (0, 1)$ , so $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2} = 1$ (property 1 below).

Non-example. $\sin 30$ is not $\frac12$ . The number 30 is a distance along the unit circle, almost five full turns, and $\sin 30 \approx -0.988$ . The angle of 30 degrees is $\frac{\pi}{6}$ radians, and $\sin\frac{\pi}{6} = \frac12$ (Example 2).

Reflecting and rotating the circle moves $P(t)$ to other points $P(\ldots)$ in a predictable way. These properties do most of the work on this page.

Remark 2 (Properties of the point

P(t)

)

Let $t$ and $\alpha$ be real numbers and $k$ an integer.

$P(0) = (1, 0)$ , $P\bigl(\frac{\pi}{2}\bigr) = (0, 1)$ , $P(\pi) = ({-1}, 0)$ and $P\bigl(\frac{3\pi}{2}\bigr) = (0, {-1})$ .
A rotation about $O$ that maps $A$ to $P(\alpha)$ maps $P(t)$ to $P(t + \alpha)$ .
$P(t + 2k\pi) = P(t)$ . Conversely, if $P(s) = P(t)$ for real numbers $s$ and $t$ , then $s - t = 2k\pi$ for some integer $k$ .
$\cos(-t) = \cos t$ and $\sin(-t) = -\sin t$ .
$\cos(\pi - t) = -\cos t$ and $\sin(\pi - t) = \sin t$ .
$\cos\bigl(\frac{\pi}{2} - t\bigr) = \sin t$ and $\sin\bigl(\frac{\pi}{2} - t\bigr) = \cos t$ .
$\cos(t + \pi) = -\cos t$ and $\sin(t + \pi) = -\sin t$ .
If $0 < t < \frac{\pi}{2}$ , then $\cos t > 0$ and $\sin t > 0$ . If $0 < t < \pi$ , then $\sin t > 0$ .

Reasons, from the facts from school above.

The axes cut the circle into four arcs, one in each quadrant. The reflection in the $y$ -axis maps the arc in the first quadrant onto the one in the second, and the reflection in the $x$ -axis maps these two onto the arcs in the fourth and third quadrants. Reflections keep lengths, so the four arcs have the same length, and as the lengths add up to $2\pi$ , each has length $\frac{\pi}{2}$ . Walking anticlockwise from $A$ we reach $(0, 1)$ , $({-1}, 0)$ and $(0, {-1})$ after $\frac{\pi}{2}$ , $\pi$ and $\frac{3\pi}{2}$ .
$P(t)$ is reached from $A$ by a journey of $t$ . The rotation keeps the length and the direction of that journey, so the image of $P(t)$ is reached from the image of $A$ , which is $P(\alpha)$ , by a journey of $t$ . Since $P(\alpha)$ is reached from $A$ by a journey of $\alpha$ , the image is reached from $A$ by a journey of $\alpha + t$ : it is $P(t + \alpha)$ .
A journey of $2\pi$ goes once round the circle and ends where it started, so $P(t + 2\pi) = P(t)$ ; repeating this $k$ times (with $t - 2\pi$ in place of $t$ when $k < 0$ ) gives $P(t + 2k\pi) = P(t)$ . Conversely, let $P(s) = P(t)$ . By the remark on the integer part of a real number in Real Numbers and Intervals, there is an integer $k$ with $k \le \frac{s - t}{2\pi} < k + 1$ . Multiplying by the positive number $2\pi$ keeps both inequalities, and subtracting $2k\pi$ keeps them too (properties 5 and 3 of the order rules of Real Numbers and Intervals), so $u = s - t - 2k\pi$ satisfies $0 \le u < 2\pi$ . Then $P(s) = P(t + u + 2k\pi) = P(t + u)$ . A rotation that maps $A$ to $P(-t)$ maps $P(t)$ to $P(0) = A$ and $P(t + u)$ to $P(u)$ , by property 2. As $P(t + u) = P(t)$ , it follows that $P(u) = A = P(0)$ . Only one length in $[0, 2\pi)$ leads from $A$ to $A$ , so $u = 0$ and $s - t = 2k\pi$ .
The reflection in the $x$ -axis maps $(x, y)$ to $(x, -y)$ , fixes $A$ and reverses directions. It maps the journey of $t$ from $A$ to a journey of $-t$ from $A$ , so it maps $P(t)$ to $P(-t)$ . Hence $(\cos(-t), \sin(-t)) = (\cos t, -\sin t)$ .
The reflection in the $y$ -axis maps $(x, y)$ to $(-x, y)$ , maps $A$ to $P(\pi) = ({-1}, 0)$ and reverses directions. It maps the journey of $t$ from $A$ to a journey of $-t$ from $P(\pi)$ , which ends at $P(\pi - t)$ . Hence $(\cos(\pi - t), \sin(\pi - t)) = (-\cos t, \sin t)$ .
The reflection in the line $y = x$ maps $(x, y)$ to $(y, x)$ , maps $A$ to $P\bigl(\frac{\pi}{2}\bigr) = (0, 1)$ and reverses directions, so as in 5 it maps $P(t)$ to $P\bigl(\frac{\pi}{2} - t\bigr)$ . Hence $\bigl(\cos\bigl(\frac{\pi}{2} - t\bigr), \sin\bigl(\frac{\pi}{2} - t\bigr)\bigr) = (\sin t, \cos t)$ .
$t + \pi = \pi - (-t)$ , so by 5 and then 4, $\cos(t + \pi) = -\cos(-t) = -\cos t$ and $\sin(t + \pi) = \sin(-t) = -\sin t$ .
A journey of $t$ with $0 < t < \frac{\pi}{2}$ ends on the arc in the first quadrant, but at neither of its ends $(1, 0)$ and $(0, 1)$ , by 1; there $x > 0$ and $y > 0$ . A journey of $t$ with $0 < t < \pi$ ends on the upper half of the circle, from $(1, 0)$ to $({-1}, 0)$ , at neither end; there $y > 0$ .

Property 4 says that $\cos$ is even and $\sin$ is odd: their common domain $\R$ is symmetric about 0. Properties 5–7 move any $t$ into the first quadrant, so the values there decide all the others.

Two more pairs of functions are built from $\sin$ and $\cos$ .

In words. $\tan t$ is the slope of the line through $O$ and $P(t)$ : the rise $\sin t$ divided by the run $\cos t$ . When $\cos t \ne 0$ , that line meets the vertical line $x = 1$ at the point $(1, \tan t)$ , which is $P(t)$ with both coordinates multiplied by $\frac{1}{\cos t}$ . The other three are reciprocals: $\sec t = \frac{1}{\cos t}$ , $\csc t = \frac{1}{\sin t}$ , and $\cot t = \frac{1}{\tan t}$ wherever both are defined. Remark 3 finds the $t$ where $\cos t$ or $\sin t$ is 0.

Example. $P(0) = (1, 0)$ gives $\tan 0 = \frac{0}{1} = 0$ and $\sec 0 = 1$ .

Non-example. $\tan\frac{\pi}{2}$ is not a number, and not “infinity” either: by property 1, $\cos\frac{\pi}{2} = 0$ , so $\frac{\pi}{2}$ is not in the domain of $\tan$ . Likewise $\cot 0$ and $\csc 0$ are undefined, because $\sin 0 = 0$ .

Main results¶

Identities and values¶

Since $P(t)$ lies on the unit circle, its coordinates satisfy the equation of the circle. This is the most used identity of trigonometry.

Where $\cos t \ne 0$ , the number $\cos^2 t$ is positive, and dividing the identity by it gives $1 + \tan^2 t = \sec^2 t$ .

The next three facts are the ones that the page The Limit of a Function uses about $\sin$ , with $t$ in radians.

Proof

We show that $\sin^2 t \le 1$ , and then that a number whose square is at most 1 lies between -1 and 1.

(a) A square is never negative, so $\cos^2 t \ge 0$ . Adding $1 - \cos^2 t$ to both sides keeps the inequality (property 3 of the order rules), so, by Theorem 1, $\sin^2 t = 1 - \cos^2 t \le 1$ . Suppose $\sin t > 1$ . Both $\sin t$ and 1 are then non-negative, and by the order part of the square-root remark (for non-negative $u$ and $v$ , $u < v$ exactly when $u^2 < v^2$ ), $1 < \sin t$ gives $1 < \sin^2 t$ , which is impossible. Suppose $\sin t < -1$ . Multiplying by the negative number ${-1}$ reverses the inequality (property 5(c) of the order rules), so $-\sin t > 1$ , and in the same way $1 < (-\sin t)^2 = \sin^2 t$ , which is impossible. So $-1 \le \sin t \le 1$ (property 7 of the order rules, negation).

(b) The same argument with $\cos$ and $\sin$ exchanged, using $\cos^2 t = 1 - \sin^2 t \le 1$ .

By Proposition 1, 1 is the largest value of $\sin$ , so these are maxima of $\sin$ . The converse of Proposition 2 holds too: $\sin$ is zero only at the multiples of $\pi$ .

Remark 3 (Where sine and cosine are zero)

Let $t$ be a real number.

$\sin t = 0$ exactly when $t = k\pi$ for an integer $k$ .
$\cos t = 0$ exactly when $t = \frac{\pi}{2} + k\pi$ for an integer $k$ .

So the natural domain of $\tan$ and $\sec$ is the set of all real numbers except $\frac{\pi}{2} + k\pi$ , $k \in \Z$ , and that of $\cot$ and $\csc$ is the set of all real numbers except $k\pi$ , $k \in \Z$ .

Reason. 1. If $t = k\pi$ , then $\sin t = 0$ by Proposition 2. Conversely, let $\sin t = 0$ . Then $\cos^2 t = 1$ by Theorem 1. If $\cos t \ge 0$ , then $\cos t$ is the non-negative square root of 1, which is 1 (the square-root remark); if $\cos t < 0$ , then $-\cos t > 0$ (multiplying by the negative number ${-1}$ reverses the inequality, by property 5(c) of the order rules), and in the same way $-\cos t = 1$ . So $P(t) = (1, 0) = P(0)$ or $P(t) = ({-1}, 0) = P(\pi)$ , by property 1 of Remark 2. By the converse in property 3, $t = 2m\pi$ or $t = \pi + 2m\pi$ for an integer $m$ : in both cases $t$ is an integer multiple of $\pi$ .

By property 6, $\cos t = \sin\bigl(\frac{\pi}{2} - t\bigr)$ . By 1, this is 0 exactly when $\frac{\pi}{2} - t = j\pi$ for an integer $j$ , that is, when $t = \frac{\pi}{2} + k\pi$ with $k = -j$ .

The addition formulas¶

How do $\sin$ and $\cos$ of a sum depend on the summands? Not by adding: $\sin(s + t)$ is not $\sin s + \sin t$ (for $s = t = \frac{\pi}{2}$ the left side is $\sin\pi = 0$ and the right side is 2). The answer mixes sines and cosines.

Proof (Rigorous track)

We compute the distance between $P(s)$ and $P(t)$ in two ways: directly, and after a rotation that moves $P(t)$ to $A$ . This gives (a), and the others follow from (a) and the properties of $P(t)$ .

(a) By property 2 of Remark 2 with $\alpha = -t$ , a rotation about $O$ that maps $A$ to $P(-t)$ maps $P(t)$ to $P(0) = A$ and $P(s)$ to $P(s - t)$ . A rotation keeps distances (the facts from school), so the distance from $P(s)$ to $P(t)$ equals the distance from $P(s - t)$ to $A$ , and so do their squares. By the distance formula and Theorem 1, the first squared distance is

\begin{aligned} &(\cos s - \cos t)^2 + (\sin s - \sin t)^2 \\ &= \cos^2 s + \sin^2 s + \cos^2 t + \sin^2 t \\ &\quad - 2\cos s \cos t - 2\sin s \sin t \\ &= 2 - 2(\cos s \cos t + \sin s \sin t), \end{aligned}

(5)

and the second is

\begin{aligned} &\bigl(\cos(s - t) - 1\bigr)^2 + \sin^2(s - t) \\ &= \cos^2(s - t) + \sin^2(s - t) \\ &\quad - 2\cos(s - t) + 1 \\ &= 2 - 2\cos(s - t). \end{aligned}

(6)

Setting them equal and solving for $\cos(s - t)$ gives (a).

(b) Apply (a) to $s$ and $-t$ ; by property 4, $\cos(-t) = \cos t$ and $\sin(-t) = -\sin t$ , so $\cos(s + t) = \cos s \cos(-t) + \sin s \sin(-t) = \cos s \cos t - \sin s \sin t$ .

(c) By property 6, $\sin(s + t) = \cos\bigl(\frac{\pi}{2} - (s + t)\bigr) = \cos\bigl(\bigl(\frac{\pi}{2} - s\bigr) - t\bigr)$ . By (a), and property 6 again ( $\cos\bigl(\frac{\pi}{2} - s\bigr) = \sin s$ and $\sin\bigl(\frac{\pi}{2} - s\bigr) = \cos s$ ),

\begin{aligned} \sin(s + t) &= \cos\Bigl(\frac{\pi}{2} - s\Bigr)\cos t \\ &\quad + \sin\Bigl(\frac{\pi}{2} - s\Bigr)\sin t \\ &= \sin s \cos t + \cos s \sin t . \end{aligned}

(7)

(d) Apply (c) to $s$ and $-t$ , and use property 4 as in (b).

(e) and (f): put $s = t$ in (c) and (b). The other two forms of (f) follow from Theorem 1: $\cos^2 t - \sin^2 t = \cos^2 t - (1 - \cos^2 t) = 2\cos^2 t - 1$ , and in the same way $= 1 - 2\sin^2 t$ .

Parts (e) and (f) are the double-angle formulas. With the values of the next section they give exact values such as $\cos\frac{\pi}{12}$ (Example 5).

Sine, angle and tangent¶

For a small positive angle $\theta$ , the height $\sin\theta$ of $P(\theta)$ , the arc $\theta$ and the height $\tan\theta$ of the point where the line $OP(\theta)$ meets the line $x = 1$ are three lengths that are almost equal. The next lemma puts them in order. It is the one fact about sine on which the limit of $\frac{\sin x}{x}$ , and so all of the calculus of sine and cosine, will rest, so its proof uses no derivatives and no limits: only areas.

A quarter of the unit circle about the origin O, with A = (1, 0) on the x-axis, the point P = (cos θ, sin θ) on the circle at angle θ, and T = (1, tan θ) directly above A on the line through O and P. Three nested regions are shaded: the triangle OAP with a dashed side AP, inside it the height sin θ of P drawn dotted; the sector OAP bounded by the arc from A to P; and the large triangle OAT, whose vertical side AT has length tan θ. — Figure 2:For $0 < \theta < \frac{\pi}{2}$ : the triangle $OAP$ (area $\frac12\sin\theta$ ) lies inside the sector $OAP$ (area $\frac{\theta}{2}$ ), which lies inside the triangle $OAT$ (area $\frac12\tan\theta$ ).

Proof

We compare the areas of three regions that lie inside one another, which gives the inequalities with $\le$ ; applying them to $\frac{\theta}{2}$ with the double-angle formulas then makes them strict.

Step 1: the inequalities with $\le$ . Let $0 < \theta < \frac{\pi}{2}$ , let $P = P(\theta) = (\cos\theta, \sin\theta)$ , and let $T = (1, \tan\theta)$ . By property 8 of Remark 2, $\cos\theta > 0$ and $\sin\theta > 0$ , so $\tan\theta = \frac{\sin\theta}{\cos\theta}$ is defined and positive (Definition 3). Consider the three regions of Figure 2.

The triangle $OAP$ has base $OA$ of length 1 on the $x$ -axis and height $\sin\theta$ , the height of $P$ above that axis. Its area is $\frac12\sin\theta$ .
The sector $OAP$ is the sector of the arc traced by the anticlockwise journey of length $\theta$ from $A$ , which ends at $P(\theta)$ ; that arc has length $\theta$ (journeys and arcs, in the facts from school). So the sector is the part of the unit disc between the radii $OA$ and $OP$ , and its area is $\frac{\theta}{2}$ .
The triangle $OAT$ has base $OA$ of length 1 and height $\tan\theta$ , because $T$ lies directly above $A$ at height $\tan\theta > 0$ . Its area is $\frac12\tan\theta$ .

They lie inside one another.

The triangle $OAP$ lies in the sector. Every point of the triangle lies on a segment from $O$ to a point $Q$ of its side $AP$ (the facts from school, sectors). The disc contains $A$ and $P$ , so it contains the segment $AP$ and with it $Q$ , and then the whole segment $OQ$ (the facts from school, distance). The region between the radii $OA$ and $OP$ contains $O$ , $A$ and $P$ (take $\lambda = 0$ or $\lambda = 1$ ), and, since $\theta < \frac{\pi}{2} < \pi$ , with any two of its points the segment between them (the facts from school, sectors); so it contains $Q$ and then the segment $OQ$ . That segment therefore lies in the part of the region in the disc, which is the sector.
The sector lies in the triangle $OAT$ . The point $T = \frac{1}{\cos\theta}\,P$ lies on the ray from $O$ through $P$ , because $\frac{1}{\cos\theta} > 0$ . So the triangle $OAT$ is the part of the region between the radii $OA$ and $OP$ that lies on the same side of the line $AT$ (the line $x = 1$ ) as $O$ , or on it: the part with $x \le 1$ (the facts from school, sectors). Every point $(x, y)$ of the disc has $x \le 1$ . Indeed, $y^2 \ge 0$ , so adding $x^2$ keeps the inequality and $x^2 \le x^2 + y^2 \le 1$ (properties 3 and 2 of the order rules). If $x > 1$ , then $x > 1 \ge 0$ , and the order part of the square-root remark would give $x^2 > 1$ , which is impossible. So the sector, the part of that region in the disc, lies in the triangle $OAT$ .

A region that lies inside another has at most its area (the facts from school), so $\frac12\sin\theta \le \frac{\theta}{2} \le \frac12\tan\theta$ . Multiplying by the positive number 2 keeps both inequalities (property 5(b) of the order rules). So, for every $\theta$ with $0 < \theta < \frac{\pi}{2}$ ,

\sin\theta \le \theta \le \tan\theta .

(9)

Step 2: the inequalities are strict. Let $0 < \theta < \frac{\pi}{2}$ again, and let $s = \sin\frac{\theta}{2}$ and $c = \cos\frac{\theta}{2}$ . Multiplying $0 < \theta < \frac{\pi}{2}$ by the positive number $\frac12$ keeps both inequalities (property 5(a) of the order rules), so $0 < \frac{\theta}{2} < \frac{\pi}{4} < \frac{\pi}{2}$ . Property 8 then gives $s > 0$ and $c > 0$ , and step 1, applied to $\frac{\theta}{2}$ , gives $s \le \frac{\theta}{2}$ and $\frac{\theta}{2} \le \tan\frac{\theta}{2} = \frac{s}{c}$ . Also $c < 1$ : $s^2 > 0$ , and adding $1 - s^2$ to both sides of $0 < s^2$ gives $1 - s^2 < 1$ (property 3 of the order rules), so $c^2 = 1 - s^2 < 1$ by Theorem 1; and $c \ge 1$ would give $c^2 \ge 1$ , by the order part of the square-root remark.

Sine. By part (e) of Theorem 2, $\sin\theta = 2sc$ . Multiplying $c < 1$ by the positive number $2s$ keeps the strict inequality, and multiplying $s \le \frac{\theta}{2}$ by the positive number 2 keeps that one (properties 5(a) and 5(b) of the order rules), so
$\sin\theta = 2sc < 2s \le 2 \cdot \frac{\theta}{2} = \theta .$
(10)
Tangent. By part (f) of Theorem 2, $\cos\theta = c^2 - s^2$ , and $\cos\theta > 0$ by property 8. Adding $c^2 - s^2$ to both sides of $0 < s^2$ gives $c^2 - s^2 < c^2$ (property 3 of the order rules), so $0 < c^2 - s^2 < c^2$ . Multiplying $c^2 - s^2 < c^2$ by the positive number $\frac{2sc}{(c^2 - s^2)\,c^2}$ keeps the strict inequality and gives $\frac{2sc}{c^2} < \frac{2sc}{c^2 - s^2}$ . Multiplying $\frac{\theta}{2} \le \frac{s}{c}$ by the positive number 2 keeps it: $2 \cdot \frac{s}{c} \ge 2 \cdot \frac{\theta}{2}$ (properties 5(a) and 5(b) of the order rules). Hence
$\begin{aligned} \tan\theta = \frac{\sin\theta}{\cos\theta} &= \frac{2sc}{c^2 - s^2} \\ &> \frac{2sc}{c^2} = 2\cdot\frac{s}{c} \\ &\ge 2 \cdot \frac{\theta}{2} = \theta . \end{aligned}$
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So $\sin\theta < \theta < \tan\theta$ .

The proof used no derivatives and no limits. It rests directly on these facts from school:

the equation of the unit circle and of the disc, and that the disc contains the segment between any two of its points;
journeys and arcs: the arc from $A$ to $P(\theta)$ traced by the journey has length $\theta$ ;
that the region between two radii whose arc is shorter than half the circle contains the segment between any two of its points, and that the triangle $OAT$ is the part of that region with $x \le 1$ ;
that every point of the triangle $OAP$ lies on a segment from $O$ to a point $Q$ of its side $AP$ ;
the area $\frac12 bh$ of a triangle, and the area $\frac{\theta}{2}$ of a sector (so the unit disc has area $\pi$ , Archimedes’ theorem);
that a region inside another has at most its area.

Through property 8, the Pythagorean identity and parts (e) and (f), it also rests on the rest of the box: the lengths of arcs, journeys along the circle, reflections and rotations. This course proves none of these facts (Real Analysis does). Two of them hide a limit: the length of an arc and the area of a sector.

Remark 4 (Why

0 < \theta < \frac{\pi}{2}

)

Each part of the hypothesis is needed.

At $\theta = 0$ all three numbers are 0, so the strict inequalities fail.
At $\theta = \frac{\pi}{2}$ , $\tan\theta$ is undefined, and beyond it the second inequality can fail: at $\theta = \pi$ , $\tan\pi = \frac{\sin\pi}{\cos\pi} = \frac{0}{-1} = 0 < \pi$ , by property 1 of Remark 2.
For $-\frac{\pi}{2} < \theta < 0$ the inequalities reverse: $\phi = -\theta$ satisfies the lemma, and by property 4 $\sin\theta = -\sin\phi$ and $\tan\theta = -\tan\phi$ , so multiplying $\sin\phi < \phi < \tan\phi$ by the negative number ${-1}$ , which reverses both inequalities (property 5(c) of the order rules), gives $\sin\theta > \theta > \tan\theta$ .

At $\theta = \frac{\pi}{4}$ the lemma gives a first bound on the number $\pi$ itself, with no decimals.

Proof

We find $\sin\frac{\pi}{4}$ and $\tan\frac{\pi}{4}$ exactly and apply Lemma 1 at $\theta = \frac{\pi}{4}$ .

Step 1: $\frac{\pi}{4}$ lies in the range of the lemma. The point $A$ of the circle is reached from $A$ by an anticlockwise journey of some length $d$ with $0 \le d < 2\pi$ (the facts from school, journeys along the circle), so $0 < 2\pi$ (property 2 of the order rules). The number $\frac18$ is positive, because $8 > 0$ (property 6 of the order rules), so multiplying $0 < 2\pi$ by $\frac18$ keeps the inequality (property 5(a) of the order rules): $0 < \frac{\pi}{4}$ . Adding $\frac{\pi}{4}$ to both sides (property 3 of the order rules) gives $\frac{\pi}{4} < \frac{\pi}{2}$ . So $0 < \frac{\pi}{4} < \frac{\pi}{2}$ .

Step 2: $\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\tan\frac{\pi}{4} = 1$ . Let $s = \sin\frac{\pi}{4}$ . By property 6 of Remark 2 with $t = \frac{\pi}{4}$ , $\cos\frac{\pi}{4} = \cos\bigl(\frac{\pi}{2} - \frac{\pi}{4}\bigr) = s$ . By Theorem 1 with $t = \frac{\pi}{4}$ , $s^2 + s^2 = 1$ , so $s^2 = \frac12$ . By property 8, $s > 0$ , since $0 < \frac{\pi}{4} < \frac{\pi}{2}$ (step 1). By the square-root remark, exactly one number $u \ge 0$ has $u^2 = \frac12$ , and $s$ is one. The number $\frac{1}{\sqrt{2}}$ is another: $\sqrt{2} \ge 0$ , and $\sqrt{2} \ne 0$ because $0^2 = 0 \ne 2$ , so $\sqrt{2} > 0$ and $\frac{1}{\sqrt{2}} > 0$ (property 6 of the order rules); and $\bigl(\frac{1}{\sqrt{2}}\bigr)^2 = \frac{1}{2}$ . So $s = \frac{1}{\sqrt{2}}$ . As $\cos\frac{\pi}{4} = s \ne 0$ , $\tan\frac{\pi}{4} = \frac{s}{s} = 1$ (Definition 3).

Step 3: the bounds. By step 1, Lemma 1 applies at $\theta = \frac{\pi}{4}$ : $\sin\frac{\pi}{4} < \frac{\pi}{4} < \tan\frac{\pi}{4}$ , that is, by step 2,

\frac{1}{\sqrt{2}} < \frac{\pi}{4} < 1 .

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Multiplying by the positive number 4 keeps both inequalities (property 5(a) of the order rules), and $\frac{4}{\sqrt{2}} = \frac{2\sqrt{2}\,\sqrt{2}}{\sqrt{2}} = 2\sqrt{2}$ , because $(\sqrt{2})^2 = 2$ (the square-root remark) and $\sqrt{2} \ne 0$ (step 2). So $2\sqrt{2} < \pi < 4$ .

In words. Doubling gives $4\sqrt{2} < 2\pi < 8$ : the circle, of length $2\pi$ , is longer than the square inscribed in it, whose perimeter is $4\sqrt{2}$ , and shorter than the square drawn around it, whose perimeter is 8; so $\pi$ lies between $2\sqrt{2} \approx 2.83$ and 4.

Graphs: period, amplitude and phase¶

As $t$ grows, $P(t)$ goes round and round the circle, and $\sin t$ and $\cos t$ repeat.

Period $2\pi$ . By property 3 of Remark 2, $2\pi$ is a period of $\sin$ and of $\cos$ (Definition 4). No smaller $p > 0$ is a period of $\sin$ : if $\sin\bigl(\frac{\pi}{2} + p\bigr) = \sin\frac{\pi}{2} = 1$ , then $\cos\bigl(\frac{\pi}{2} + p\bigr) = 0$ by Theorem 1, so $P\bigl(\frac{\pi}{2} + p\bigr) = (0, 1) = P\bigl(\frac{\pi}{2}\bigr)$ , and $p = 2k\pi$ for an integer $k$ by the converse in property 3. Multiplying $2k\pi = p > 0$ by the positive number $\frac{1}{2\pi}$ gives $k > 0$ (properties 6 and 5(a) of the order rules), so $k \ge 1$ , as $k$ is an integer; and multiplying $k \ge 1$ by the positive number $2\pi$ gives $p \ge 2\pi$ (property 5(b) of the order rules). The same argument at $t = 0$ shows it for $\cos$ . So $2\pi$ is the period of $\sin$ and of $\cos$ .
Values. The values of $\sin$ and $\cos$ fill the interval $[-1, 1]$ : they lie in it by Proposition 1, and every $y \in [-1, 1]$ is a value of $\sin$ . Indeed $y^2 \le 1$ (for $y \ge 0$ by the order part of the square-root remark, and for $y < 0$ by the same applied to $-y$ , which lies in $(0, 1]$ because multiplying $-1 \le y < 0$ by the negative number ${-1}$ reverses both inequalities, property 5(c) of the order rules), so $1 - y^2 \ge 0$ (adding $-y^2$ , property 3 of the order rules) and the point $\bigl(\sqrt{1 - y^2}, y\bigr)$ lies on the unit circle. By the facts from school (journeys along the circle) it is $P(t)$ for some $t$ ; then $\sin t = y$ . In the same way, the point $\bigl(y, \sqrt{1 - y^2}\bigr)$ shows that $y$ is a value of $\cos$ .
Shape. On $[0, 2\pi]$ the graph of $\sin$ takes the values 0, 1, 0, ${-1}$ and 0 at 0, $\frac{\pi}{2}$ , $\pi$ , $\frac{3\pi}{2}$ and $2\pi$ (property 1), and then repeats; in between, the widget below shows a smooth wave. By property 6, $\cos t = \sin\bigl(t + \frac{\pi}{2}\bigr)$ (apply 6 to $-t$ and use 4: $\sin\bigl(\frac{\pi}{2} - (-t)\bigr) = \cos(-t) = \cos t$ ), so the graph of $\cos$ is the graph of $\sin$ shifted $\frac{\pi}{2}$ to the left.

Stretching and shifting the graph of $\sin$ gives every wave of the same shape. For numbers $a > 0$ , $b > 0$ , $c$ and $d$ , the function

f(x) = a \sin\bigl(b(x - c)\bigr) + d

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has

amplitude $a$ : its values fill $[d - a, d + a]$ , because those of $\sin$ fill $[-1, 1]$ , and multiplying $-1 \le u \le 1$ by the positive number $a$ and then adding $d$ keeps both inequalities (properties 5(b) and 3 of the order rules);
period $\frac{2\pi}{b}$ : shifting $x$ by $p$ shifts $b(x - c)$ by $bp$ , and as $b(x - c)$ runs through all real numbers when $x$ does, $p$ is a period of $f$ exactly when $bp$ is a period of $\sin$ ; the smallest such $bp$ is $2\pi$ ;
phase shift $c$ : its graph is that of $a\sin(bx) + d$ shifted $c$ to the right;
midline $y = d$ : the wave swings between $d - a$ and $d + a$ , about the line $y = d$ .

Figure 3:Graph of $y = a\sin\bigl(b(x - c)\bigr) + d$ for $-7 \le x \le 7$ , with four sliders: $a$ from 0.5 to 3 in steps of 0.5, $b$ from 0.5 to 4 in steps of 0.5, $c$ from ${-3}$ to 3 in steps of 0.25, and $d$ from ${-2}$ to 2 in steps of 0.5. It starts at $a = 1$ , $b = 1$ , $c = 0$ and $d = 0$ , with the graph of $y = \sin x$ : a wave between ${-1}$ and 1 that crosses the $x$ -axis at the multiples of $\pi \approx 3.14$ and repeats every $2\pi \approx 6.28$ . Increasing $a$ stretches the wave vertically, increasing $b$ squeezes it horizontally, increasing $c$ shifts it to the right, and increasing $d$ shifts it up.

Try this:

Set $b = 2$ . How many complete waves fit between 0 and $2\pi \approx 6.28$ now? What is the period?
Set $b = 1$ again and move $c$ to ${-1.5}$ , the slider’s closest value to $-\frac{\pi}{2} \approx -1.571$ . The graph is now $y = \sin(x + 1.5)$ . Which function from this page does it almost match, and which value of $c$ , not on the slider, would match it exactly? (Look at the shape bullet above.)
Set $c = 0$ , $a = 3$ and $d = 2$ . Between which heights does the wave swing, and where is its midline?

Worked examples¶

Example 1 (Degrees, arcs and sectors on a wheel)

A wheel of radius 0.3 metres turns through 150 degrees. Find the angle in radians, the distance travelled by a point on the rim, and the area swept out by a spoke.

Radians. A full turn is 360 degrees and $2\pi$ radians, so 1 degree is $\frac{\pi}{180}$ radians, and
$150 \cdot \frac{\pi}{180} = \frac{5\pi}{6} .$
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Arc. On a circle of radius $r$ , an angle of $\theta$ radians cuts off an arc of length $r\theta$ (Radian and the facts on circles of other radii). Here $0.3 \cdot \frac{5\pi}{6} = \frac{\pi}{4}$ , in metres.
Sector. The sector has area $\frac12 r^2\theta = \frac12 \cdot 0.09 \cdot \frac{5\pi}{6} = \frac{3\pi}{80}$ , in square metres.

\boxed{ \begin{aligned} \theta &= \frac{5\pi}{6}, \\ r\theta &= \frac{\pi}{4} \approx 0.785, \\ \tfrac12 r^2\theta &= \frac{3\pi}{80} \approx 0.118 \end{aligned} }

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The arc is in metres and the area in square metres.

Check. 150 degrees is $\frac{150}{360} = \frac{5}{12}$ of a full turn. The circumference is $2\pi \cdot 0.3 = 0.6\pi$ , and $\frac{5}{12} \cdot 0.6\pi = \frac{\pi}{4}$ ✓. The disc has area $\pi \cdot 0.09$ , and $\frac{5}{12} \cdot 0.09\pi = 0.0375\pi = \frac{3\pi}{80}$ ✓.

Example 2 (Exact values at

\frac{\pi}{4}

\frac{\pi}{3}

and

\frac{\pi}{6}

)

Find $\cos t$ and $\sin t$ exactly for $t = \frac{\pi}{4}$ , $\frac{\pi}{3}$ and $\frac{\pi}{6}$ .

At $\frac{\pi}{4}$ . By property 6 of Remark 2 with $t = \frac{\pi}{4}$ , $\cos\frac{\pi}{4} = \sin\bigl(\frac{\pi}{2} - \frac{\pi}{4}\bigr)$ , which is $\sin\frac{\pi}{4}$ ; call this common value $c$ . By Theorem 1, $c^2 + c^2 = 1$ , so $c^2 = \frac12$ . By property 8, $c > 0$ , so $c$ is the non-negative square root of $\frac12$ (the square-root remark). Since $\frac{\sqrt{2}}{2} \ge 0$ and $\bigl(\frac{\sqrt{2}}{2}\bigr)^2 = \frac{2}{4} = \frac12$ , that root is $\frac{\sqrt{2}}{2}$ .
At $\frac{\pi}{3}$ . Let $c = \cos\frac{\pi}{3}$ . Part (f) of Theorem 2 gives $\cos\frac{2\pi}{3} = 2c^2 - 1$ , and property 5 gives $\cos\frac{2\pi}{3} = \cos\bigl(\pi - \frac{\pi}{3}\bigr) = -c$ . So $2c^2 - 1 = -c$ , that is, $2c^2 + c - 1 = 0$ , which factors as $(2c - 1)(c + 1) = 0$ . By property 8, $c > 0$ , so adding 1 gives $c + 1 > 1 > 0$ (properties 3 and 2 of the order rules; $1 > 0$ because multiplying $0 < \frac{1}{c}$ by $c > 0$ gives $0 < 1$ , by properties 6 and 5(a)), and therefore $2c - 1 = 0$ : $c = \frac12$ . Then $\sin^2\frac{\pi}{3} = 1 - \frac14 = \frac34$ , and $\sin\frac{\pi}{3} > 0$ , so, as in step 1, $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ .
At $\frac{\pi}{6}$ . Since $\frac{\pi}{6} = \frac{\pi}{2} - \frac{\pi}{3}$ , property 6 with $t = \frac{\pi}{3}$ gives $\cos\frac{\pi}{6} = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \cos\frac{\pi}{3} = \frac12$ .

So:

$\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac12$ ;
$\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ ;
$\cos\frac{\pi}{3} = \frac12$ and $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ .

Dividing, $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ , $\tan\frac{\pi}{4} = 1$ and $\tan\frac{\pi}{3} = \sqrt{3}$ .

Check. With a calculator in radian mode, $\frac{\pi}{3} \approx 1.047198$ gives $\cos 1.047198 \approx 0.500000$ ✓, and $\frac{\pi}{4} \approx 0.785398$ gives $\sin 0.785398 \approx 0.707107 \approx \frac{1.414214}{2}$ ✓. Each pair satisfies $\cos^2 t + \sin^2 t = 1$ : for example $\frac14 + \frac34 = 1$ ✓.

Example 3 (Angles outside the first quadrant)

Find $\sin\frac{5\pi}{6}$ , $\cos\frac{4\pi}{3}$ , $\tan\bigl(-\frac{\pi}{4}\bigr)$ and $\sin\frac{17\pi}{6}$ .

The properties of Remark 2 move each number to the first quadrant, where Example 2 gives the values.

$\frac{5\pi}{6} = \pi - \frac{\pi}{6}$ , so by property 5, $\sin\frac{5\pi}{6} = \sin\frac{\pi}{6} = \frac12$ .
$\frac{4\pi}{3} = \frac{\pi}{3} + \pi$ , so by property 7, $\cos\frac{4\pi}{3} = -\cos\frac{\pi}{3} = -\frac12$ .
By property 4, $\sin\bigl(-\frac{\pi}{4}\bigr) = -\sin\frac{\pi}{4}$ and $\cos\bigl(-\frac{\pi}{4}\bigr) = \cos\frac{\pi}{4} \ne 0$ , so $\tan\bigl(-\frac{\pi}{4}\bigr) = \frac{-\sin(\pi/4)}{\cos(\pi/4)} = -\tan\frac{\pi}{4} = -1$ .
$\frac{17\pi}{6} = \frac{5\pi}{6} + 2\pi$ , so by property 3 and step 1, $\sin\frac{17\pi}{6} = \sin\frac{5\pi}{6} = \frac12$ .

\boxed{ \begin{aligned} \sin\frac{5\pi}{6} &= \frac12, \\ \cos\frac{4\pi}{3} &= -\frac12, \\ \tan\Bigl(-\frac{\pi}{4}\Bigr) &= -1, \\ \sin\frac{17\pi}{6} &= \frac12 \end{aligned} }

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Check. The signs agree with the quadrants: $P\bigl(\frac{5\pi}{6}\bigr)$ is in the second quadrant, where $y > 0$ ; $P\bigl(\frac{4\pi}{3}\bigr)$ in the third, where $x < 0$ ; $P\bigl(-\frac{\pi}{4}\bigr)$ in the fourth, where $x$ and $y$ have opposite signs ✓. With a calculator in radian mode, $\sin 8.901179 \approx 0.500000$ , and $\frac{17\pi}{6} \approx 8.901179$ ✓.

Example 4 (The cosine from the sine, and the sign)

Given that $\sin t = \frac35$ and $\frac{\pi}{2} < t < \pi$ , find $\cos t$ and $\tan t$ .

The square. By Theorem 1, $\cos^2 t = 1 - \frac{9}{25} = \frac{16}{25}$ .
The sign. Let $u = \pi - t$ . Multiplying $\frac{\pi}{2} < t < \pi$ by the negative number ${-1}$ reverses both inequalities, and adding $\pi$ keeps them (properties 5(c) and 3 of the order rules): $0 < u < \frac{\pi}{2}$ . By property 5 of Remark 2, $\cos t = \cos(\pi - u) = -\cos u$ , and $\cos u > 0$ by property 8. So $\cos t < 0$ .
The value. So $-\cos t$ is the non-negative number whose square is $\frac{16}{25}$ , which is $\frac45$ (the square-root remark): $\cos t = -\frac45$ .
The tangent. $\tan t = \dfrac{3/5}{-4/5} = -\dfrac34$ .

\boxed{\cos t = -\frac45, \qquad \tan t = -\frac34}

(18)

Check. $\bigl(\frac35\bigr)^2 + \bigl(-\frac45\bigr)^2 = \frac{9 + 16}{25} = 1$ ✓. With a calculator in radian mode, $t = 2.498092$ lies between $\frac{\pi}{2}$ and $\pi$ and gives $\sin t \approx 0.600000$ and $\cos t \approx -0.800000$ ✓.

Example 5 (An exact value from the addition formulas)

Find $\cos\frac{\pi}{12}$ exactly.

Write the angle as a difference of known ones. $\frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{\pi}{12}$ .
Apply part (a) of Theorem 2, with the values of Example 2:
$\begin{aligned} \cos\frac{\pi}{12} &= \cos\frac{\pi}{3}\cos\frac{\pi}{4} \\ &\quad + \sin\frac{\pi}{3}\sin\frac{\pi}{4} \\ &= \frac12 \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2} + \sqrt{6}}{4}, \end{aligned}$
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since $\sqrt{3}\,\sqrt{2} = \sqrt{6}$ (both sides are non-negative and square to 6).

\boxed{\cos\frac{\pi}{12} = \frac{\sqrt{2} + \sqrt{6}}{4}}

(20)

Check. $\frac{1.414214 + 2.449490}{4} \approx 0.965926$ , and with a calculator in radian mode $\cos 0.261799 \approx 0.965926$ , where $\frac{\pi}{12} \approx 0.261799$ ✓.

Example 6 (The Ferris wheel)

The Ferris wheel of Why this matters has radius 20 metres and its centre 22 metres above the ground, turns anticlockwise as we look at it, once every 10 minutes, and you board at the bottom at time 0. Find your height $h(t)$ , in metres, after $t$ minutes, its amplitude, period, phase shift and midline, and your height after 100 seconds and after 25 minutes.

The angle. The wheel turns $2\pi$ radians in 10 minutes at a steady speed, so after $t$ minutes it has turned through $\frac{2\pi t}{10} = \frac{\pi t}{5}$ radians.
The position. The wheel is the unit circle scaled by 20 and moved to the centre (the facts on circles of other radii). You start at the bottom, which corresponds to $P\bigl(-\frac{\pi}{2}\bigr) = (0, {-1})$ (properties 4 and 1 of Remark 2). Turning through $\theta$ is a rotation that maps $A$ to $P(\theta)$ , so by property 2 it takes you to the point corresponding to $P\bigl(\theta - \frac{\pi}{2}\bigr)$ , at height $20\sin\bigl(\theta - \frac{\pi}{2}\bigr)$ above the centre.
Simplify. By property 4 and then property 6, $\sin\bigl(\theta - \frac{\pi}{2}\bigr) = -\sin\bigl(\frac{\pi}{2} - \theta\bigr) = -\cos\theta$ . With $\theta = \frac{\pi t}{5}$ :
$h(t) = 22 - 20\cos\frac{\pi t}{5} .$
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Amplitude, period, phase shift, midline. Step 3 also says that $-\cos\theta = \sin\bigl(\theta - \frac{\pi}{2}\bigr)$ , and $\frac{\pi t}{5} - \frac{\pi}{2} = \frac{\pi}{5}\bigl(t - \frac52\bigr)$ , so $h(t) = 20\sin\bigl(\frac{\pi}{5}\bigl(t - \frac52\bigr)\bigr) + 22$ : amplitude 20, period $\frac{2\pi}{\pi/5} = 10$ , phase shift $\frac52$ and midline $h = 22$ . The heights fill $[22 - 20, 22 + 20] = [2, 42]$ .
The two times. 100 seconds is $\frac53$ minutes, and $\frac{\pi}{5}\cdot\frac53 = \frac{\pi}{3}$ , so $h\bigl(\frac53\bigr) = 22 - 20 \cdot \frac12 = 12$ (Example 2). After 25 minutes, $\cos(5\pi) = \cos(\pi + 4\pi) = \cos\pi = -1$ (properties 3 and 1), so $h(25) = 22 + 20 = 42$ : you are at the top.

\boxed{ \begin{aligned} h(t) &= 22 - 20\cos\frac{\pi t}{5}, \\ h\Bigl(\frac53\Bigr) &= 12, \qquad h(25) = 42 \end{aligned} }

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The heights are in metres and the times in minutes.

Check. $h(0) = 22 - 20 = 2$ : you board at the bottom ✓. $h(10) = 22 - 20\cos 2\pi = 2$ : one turn later you are back ✓. Halfway round, $h(5) = 22 - 20\cos\pi = 42$ , the top ✓.

Common mistakes¶

Rigorous track¶

Why radians, and not degrees?

In degrees, the angle that cuts off an arc of length $\theta$ is $\frac{180\theta}{\pi}$ , so the “sine in degrees” is $s(x) = \sin\frac{\pi x}{180}$ . Then Lemma 1 would read $s(x) < \frac{\pi x}{180} < \tan\frac{\pi x}{180}$ for $0 < x < 90$ : the sine is no longer compared with the angle $x$ itself, but with $\frac{\pi}{180}$ times it. In the same way, with $\theta = \frac{\pi x}{180}$ ,

\frac{s(x)}{x} = \frac{\pi}{180} \cdot \frac{\sin\theta}{\theta},

(23)

so every statement about $\frac{\sin\theta}{\theta}$ carries the factor $\frac{\pi}{180}$ when it is written in degrees. Radians are the unit in which the length of the arc is the angle, and no such factor appears.

Defining sine without pictures

Definition 2 rests on the length of an arc, and Lemma 1 on the area of a sector. Both are limits in disguise (the length of a curve is a limit of the lengths of inscribed polygons), so a fully rigorous treatment must define them first. There are two standard ways round this. One defines $\sin$ and $\cos$ by power series and proves every property on this page from the series; the number $\pi$ is then defined as twice the smallest positive zero of $\cos$ . The other defines the area of a sector as an integral and the angle from it. Either way, the facts from school become theorems. The course takes them as known here, and the order of the pages makes sure that they are never “proved” by a result that rests on them; that construction belongs to Real Analysis (ana).

Summary¶

An angle in radians is the length of the arc it cuts off on the unit circle: a full turn is $2\pi$ , and 180 degrees is $\pi$ . On a circle of radius $r$ , the arc is $r\theta$ and the sector $\frac12 r^2\theta$ .
$P(t) = (\cos t, \sin t)$ is the point reached by walking a distance $t$ from $(1, 0)$ round the unit circle, anticlockwise for $t > 0$ (Definition 2); $\tan t = \frac{\sin t}{\cos t}$ .
Reflections and rotations of the circle give the symmetries (Remark 2): period $2\pi$ , $\cos$ even, $\sin$ odd, and values in any quadrant from values in the first.
$\cos^2 t + \sin^2 t = 1$ ; $-1 \le \sin t \le 1$ ; $\sin(k\pi) = 0$ and $\sin\bigl(\frac{\pi}{2} + 2k\pi\bigr) = 1$ for every integer $k$ .
Addition formulas: $\sin(s + t) = \sin s\cos t + \cos s\sin t$ and $\cos(s + t) = \cos s\cos t - \sin s\sin t$ .
For $0 < \theta < \frac{\pi}{2}$ : $\sin\theta < \theta < \tan\theta$ , by comparing areas, with no derivatives (Lemma 1).
$a\sin\bigl(b(x - c)\bigr) + d$ (with $a, b > 0$ ) has amplitude $a$ , period $\frac{2\pi}{b}$ , phase shift $c$ and midline $y = d$ .

Exercises¶

Exercise 12 (Squeezing

\frac{\sin\theta}{\theta}

)

Use Lemma 1 to show that

\cos\theta < \frac{\sin\theta}{\theta} < 1

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for every $\theta$ with $0 < \theta < \frac{\pi}{2}$ , and then for every $\theta$ with $-\frac{\pi}{2} < \theta < 0$ .

Hint 1

Divide $\sin\theta < \theta$ by $\theta$ , and multiply $\theta < \tan\theta$ by $\cos\theta$ . What are the signs of $\theta$ and $\cos\theta$ ?

Hint 2

For $\theta < 0$ , put $\phi = -\theta$ and use property 4 of Remark 2.

Answer

From $\sin\theta < \theta < \frac{\sin\theta}{\cos\theta}$ , multiplying by the positive numbers $\frac{1}{\theta}$ and $\frac{\cos\theta}{\theta}$ ; for $\theta < 0$ , both sides are unchanged when $\theta$ is replaced by $-\theta$ .

Where this leads¶

These topics build on this one:

The Squeeze Theorem
Inverse Trigonometric Functions (coming soon)
Complex Numbers (coming soon)

Found an error on this page? Report it (the form opens with this page filled in).

The page The Squeeze Theorem traps $\frac{\sin x}{x}$ between $\cos x$ and 1, as in Exercise 12, and finds its limit at 0; from it come the derivatives of all six trigonometric functions.