High School Mathematics · Grades 10–12

14Trigonometry: The Unit Circle

Trigonometry begins in right triangles, but its true home is the unit circle, where cosine and sine become functions of an arbitrary real number. This chapter installs the radian, the winding of the real line around the circle, the exact values to know by heart, and the symmetries that generate all the classical identities. The study of $\cos$ and $\sin$ as functions — variations, derivatives — is carried out in Chapter 24.

14.1 Radians and the winding

Definition 14.1 (Radian)

Let $\mathcal C$ be the circle of radius $1$ centered at the origin $O$ , and $I(1, 0)$ . The radian measure of an angle at the center of $\mathcal C$ is the length of the arc it cuts on $\mathcal C$ . Since the full circle has circumference $2\pi$ :

360^\circ = 2\pi \text{ rad}, \qquad 180^\circ = \pi \text{ rad}, \qquad 1 \text{ rad} = \frac{180^\circ}{\pi} \approx 57.3^\circ .

Method 14.2 (Converting)

Degrees and radians are proportional: multiply by $\frac{\pi}{180}$ to go from degrees to radians, by $\frac{180}{\pi}$ the other way. For instance $60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3}$ , and $\frac{3\pi}{4} = \frac{3 \times 180^\circ}{4} = 135^\circ$ .

Definition 14.3 (Winding the line on the circle)

To each real number $t$ , associate the point $M(t)$ of $\mathcal C$ obtained by walking a distance $\abs t$ along the circle from $I$ , counterclockwise if $t \geq 0$ , clockwise if $t < 0$ . Since the circumference is $2\pi$ , the numbers $t$ and $t + 2k\pi$ ( $k \in \Z$ ) reach the same point: $M(t + 2k\pi) = M(t)$ .

Example 14.4

$M(0) = I$ ; $M\!\left(\frac{\pi}{2}\right) = (0, 1)$ , the top of the circle; $M(\pi) = (-1, 0)$ ; $M\!\left(-\frac{\pi}{2}\right) = (0, -1)$ ; and $M\!\left(\frac{9\pi}{4}\right) = M\!\left(\frac{\pi}{4} + 2\pi\right) = M\!\left(\frac{\pi}{4}\right)$ .

14.2 Cosine and sine

Definition 14.5 (Cosine and sine)

For every real $t$ , the cosine and sine of $t$ are the coordinates of the point $M(t)$ of the unit circle:

M(t) = (\cos t,\ \sin t).

The unit circle: the real number t, wound counterclockwise from I, lands at the point M(t) whose coordinates define t and t. — The unit circle: the real number $t$ , wound counterclockwise from $I$ , lands at the point $M(t)$ whose coordinates *define* $\cos t$ and $\sin t$ .

Proposition 14.6 (First properties)

For all $t \in \R$ and $k \in \Z$ :

-1 \leq \cos t \leq 1, \qquad -1 \leq \sin t \leq 1, \qquad \cos^2 t + \sin^2 t = 1,

\cos(t + 2k\pi) = \cos t, \qquad \sin(t + 2k\pi) = \sin t .

Proof. $M(t)$ lies on the unit circle, so its coordinates are between $-1$ and $1$ , and they satisfy the circle’s equation $x^2 + y^2 = 1$ , which is the identity $\cos^2 t + \sin^2 t = 1$ . Periodicity restates $M(t + 2k\pi) = M(t)$ (Definition 14.3). ∎

Example 14.7

If $\sin t = \frac35$ and $t \in \intcc{\frac\pi2}{\pi}$ (second quadrant), then $\cos^2 t = 1 - \frac{9}{25} = \frac{16}{25}$ , so $\cos t = \pm\frac45$ ; in the second quadrant the abscissa is negative, hence $\cos t = -\frac45$ .

The values to know by heart:

$t$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$
[6pt] $\cos t$	$1$	$\dfrac{\sqrt3}{2}$	$\dfrac{\sqrt2}{2}$	$\dfrac12$	$0$
[6pt] $\sin t$	$0$	$\dfrac12$	$\dfrac{\sqrt2}{2}$	$\dfrac{\sqrt3}{2}$	$1$

Remark 14.8

A memory aid: the sines read $\frac{\sqrt0}{2}, \frac{\sqrt1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2}, \frac{\sqrt4}{2}$ , and the cosines are the same list reversed.

14.3 Associated angles

Proposition 14.9 (Associated angles)

For all $t \in \R$ :

\begin{aligned} \cos(-t) &= \cos t, & \sin(-t) &= -\sin t,\\ \cos(\pi - t) &= -\cos t, & \sin(\pi - t) &= \sin t,\\ \cos(\pi + t) &= -\cos t, & \sin(\pi + t) &= -\sin t,\\ \cos\left(\tfrac{\pi}{2} - t\right) &= \sin t, & \sin\left(\tfrac{\pi}{2} - t\right) &= \cos t . \end{aligned}

Proof. Each line expresses a symmetry of the circle.

$M(-t)$ is the reflection of $M(t)$ in the $x$ -axis: the abscissa is unchanged, the ordinate changes sign.

$M(\pi - t)$ is the reflection of $M(t)$ in the $y$ -axis: walking backwards from $\pi$ by $t$ lands opposite (horizontally) to walking forwards from $0$ by $t$ . The abscissa changes sign, the ordinate is unchanged.

$M(\pi + t)$ is diametrically opposite $M(t)$ (half a turn): both coordinates change sign.

$M\!\left(\frac\pi2 - t\right)$ is the reflection of $M(t)$ in the diagonal line $y = x$ , which swaps the two coordinates. ∎

The four associated points: M(-t) mirrors M(t) in the x-axis, M(π - t) in the y-axis, and M(π + t) is diametrically opposite. Every identity of is read off this picture. — The four associated points: $M(-t)$ mirrors $M(t)$ in the $x$ -axis, $M(\pi - t)$ in the $y$ -axis, and $M(\pi + t)$ is diametrically opposite. Every identity of Proposition 14.9 is read off this picture.

Example 14.10

$\cos\frac{2\pi}{3} = \cos\left(\pi - \frac\pi3\right) = -\cos\frac\pi3 = -\frac12$ , and $\sin\frac{7\pi}{6} = \sin\left(\pi + \frac\pi6\right) = -\sin\frac\pi6 = -\frac12$ . The table of five values, extended by the symmetries, covers the whole circle.

14.4 Solving trigonometric equations

Theorem 14.11 (The equations $\cos t = \cos a$ and $\sin t = \sin a$ )

For real numbers $t$ and $a$ :

\cos t = \cos a \iff t = a + 2k\pi \ \text{or}\ t = -a + 2k\pi \quad (k \in \Z),

\sin t = \sin a \iff t = a + 2k\pi \ \text{or}\ t = \pi - a + 2k\pi \quad (k \in \Z).

Proof. Two points of the unit circle have the same abscissa exactly when they are equal or reflections of each other in the $x$ -axis; by Proposition 14.9, the reflection of $M(a)$ is $M(-a)$ . So $\cos t = \cos a$ means $M(t) = M(a)$ or $M(t) = M(-a)$ , i.e. $t = \pm a$ up to a multiple of $2\pi$ . Similarly, equal ordinates mean equal points or reflections in the $y$ -axis, and the reflection of $M(a)$ is $M(\pi - a)$ . ∎

Method 14.12 (Solving $\cos t = c$ on an interval)

Find one angle $a$ with $\cos a = c$ , from the table of known values.
Write the general solutions $t = \pm a + 2k\pi$ (Theorem 14.11).
Choose the integers $k$ that land inside the requested interval, and list the solutions.

Proceed likewise for $\sin t = c$ with $t = a + 2k\pi$ or $t = \pi - a + 2k\pi$ .

Example 14.13

Solve $\cos t = \frac12$ on $\intoc{-\pi}{\pi}$ . One solution of $\cos a = \frac12$ is $a = \frac{\pi}{3}$ . The general solutions are $t = \frac\pi3 + 2k\pi$ and $t = -\frac\pi3 + 2k\pi$ . Inside $\intoc{-\pi}{\pi}$ , only $k = 0$ contributes: $t \in \left\{-\frac\pi3,\ \frac\pi3\right\}$ .

Solve $\sin t = -\frac{\sqrt2}{2}$ on $\intco{0}{2\pi}$ . One angle is $a = -\frac\pi4$ ; the solutions are $t = -\frac\pi4 + 2k\pi$ and $t = \pi + \frac\pi4 + 2k\pi = \frac{5\pi}{4} + 2k\pi$ . In $\intco{0}{2\pi}$ : $t = \frac{7\pi}{4}$ (from $k = 1$ ) and $t = \frac{5\pi}{4}$ .

Solving t = 1/2: the vertical line x = 1/2 (orange) cuts the unit circle in two points, symmetric in the x-axis — hence the two solution families t = ± π3 + 2kπ. — Solving $\cos t = \frac12$ : the vertical line $x = \frac12$ (orange) cuts the unit circle in two points, symmetric in the $x$ -axis — hence the two solution families $t = \pm\frac\pi3 + 2k\pi$ .

14.5 Exercises

Exercise 14.1 ★

Convert to radians: $30^\circ$ , $45^\circ$ , $120^\circ$ , $270^\circ$ . Convert to degrees: $\frac{\pi}{5}$ , $\frac{5\pi}{6}$ , $\frac{7\pi}{4}$ , $\frac{2\pi}{9}$ .

Solution

Solution of Exercise 14.1.

To radians: $30^\circ = \frac{\pi}{6}$ , $45^\circ = \frac{\pi}{4}$ , $120^\circ = \frac{2\pi}{3}$ , $270^\circ = \frac{3\pi}{2}$ .

To degrees: $\frac{\pi}{5} = 36^\circ$ , $\frac{5\pi}{6} = 150^\circ$ , $\frac{7\pi}{4} = 315^\circ$ , $\frac{2\pi}{9} = 40^\circ$ .

Exercise 14.2 ★

Place on the unit circle the points $M(t)$ for

t = \frac{2\pi}{3},\quad t = -\frac{\pi}{4},\quad t = \frac{17\pi}{6},\quad t = -\frac{7\pi}{2},

after reducing each to a value in $\intoc{-\pi}{\pi}$ modulo $2\pi$ .

Solution

Solution of Exercise 14.2.

$\frac{2\pi}{3}$ and $-\frac{\pi}{4}$ are already in $\intoc{-\pi}{\pi}$ : second quadrant and fourth quadrant respectively.

$\frac{17\pi}{6} - 2\pi = \frac{5\pi}{6}$ : second quadrant, close to the negative $x$ -axis.

$-\frac{7\pi}{2} + 4\pi = \frac{\pi}{2}$ : the top of the circle, $(0, 1)$ .

Exercise 14.3 ★

Using the table and the associated angles, give the exact values of

\cos\frac{3\pi}{4}, \quad \sin\frac{5\pi}{6}, \quad \cos\left(-\frac{\pi}{3}\right), \quad \sin\frac{4\pi}{3}, \quad \cos\frac{11\pi}{6} .

Solution

Solution of Exercise 14.3.

$\cos\frac{3\pi}{4} = \cos\left(\pi - \frac\pi4\right) = -\cos\frac\pi4 = -\frac{\sqrt2}{2}$ .

$\sin\frac{5\pi}{6} = \sin\left(\pi - \frac\pi6\right) = \sin\frac\pi6 = \frac12$ .

$\cos\left(-\frac\pi3\right) = \cos\frac\pi3 = \frac12$ .

$\sin\frac{4\pi}{3} = \sin\left(\pi + \frac\pi3\right) = -\sin\frac\pi3 = -\frac{\sqrt3}{2}$ .

$\cos\frac{11\pi}{6} = \cos\left(-\frac\pi6 + 2\pi\right) = \cos\frac\pi6 = \frac{\sqrt3}{2}$ .

Exercise 14.4 ★

Given $\cos t = \frac{5}{13}$ with $t \in \intoo{-\frac\pi2}{0}$ , compute $\sin t$ exactly.

Solution

Solution of Exercise 14.4.

$\sin^2 t = 1 - \cos^2 t = 1 - \frac{25}{169} = \frac{144}{169}$ , so $\sin t = \pm\frac{12}{13}$ . On $\intoo{-\frac\pi2}{0}$ (fourth quadrant) the ordinate is negative: $\sin t = -\frac{12}{13}$ .

Exercise 14.5 ★★

Simplify, for any real $x$ :

A = \cos(\pi + x) + \cos(-x) + \sin\left(\frac\pi2 - x\right) - \cos x, \qquad B = \sin(\pi - x) + \sin(\pi + x) + \sin(-x) .

Solution

Solution of Exercise 14.5.

Using Proposition 14.9:

A = (-\cos x) + \cos x + \cos x - \cos x = 0 ,

B = \sin x + (-\sin x) + (-\sin x) = -\sin x .

Exercise 14.6 ★★

Solve on $\intoc{-\pi}{\pi}$ :

\cos t = \frac{\sqrt3}{2}, \qquad \sin t = \frac12, \qquad \cos t = -1 .

Solution

Solution of Exercise 14.6.

$\cos t = \frac{\sqrt3}{2}$ : one solution is $\frac\pi6$ , so $t = \pm\frac\pi6$ (the shifts by $2k\pi$ leave the interval).

$\sin t = \frac12$ : $t = \frac\pi6$ or $t = \pi - \frac\pi6 = \frac{5\pi}{6}$ .

$\cos t = -1$ : the only point of abscissa $-1$ is $M(\pi)$ , so $t = \pi$ .

Exercise 14.7 ★★

Solve on $\intco{0}{2\pi}$ :

\sin t = -\frac{\sqrt3}{2}, \qquad \cos t = 0, \qquad 2\sin t + 1 = 0 .

Solution

Solution of Exercise 14.7.

$\sin t = -\frac{\sqrt3}{2}$ : from $a = -\frac\pi3$ , the families are $-\frac\pi3 + 2k\pi$ and $\pi + \frac\pi3 + 2k\pi$ ; in $\intco{0}{2\pi}$ : $t = \frac{4\pi}{3}$ and $t = \frac{5\pi}{3}$ .

$\cos t = 0$ : $t = \frac\pi2$ and $t = \frac{3\pi}{2}$ .

$2\sin t + 1 = 0$ means $\sin t = -\frac12$ : from $a = -\frac\pi6$ , in $\intco{0}{2\pi}$ : $t = \frac{7\pi}{6}$ and $t = \frac{11\pi}{6}$ .

Exercise 14.8 ★★

Solve on $\R$ the equation $\cos 2t = \cos t$ . (Use Theorem 14.11 with $a = t$ , and discuss the two families of solutions.)

Solution

Solution of Exercise 14.8.

By Theorem 14.11, $\cos 2t = \cos t$ means $2t = t + 2k\pi$ or $2t = -t + 2k\pi$ , i.e. $t = 2k\pi$ or $t = \frac{2k\pi}{3}$ ( $k \in \Z$ ). The first family is contained in the second (take $k$ a multiple of $3$ ), so the solution set is

S = \left\{\frac{2k\pi}{3} : k \in \Z\right\}

— on the circle, the three points $M(0)$ , $M\!\left(\frac{2\pi}{3}\right)$ , $M\!\left(\frac{4\pi}{3}\right)$ .

Exercise 14.9 ★★

Show that for all real $x$ ,

\bigl(\cos x + \sin x\bigr)^2 + \bigl(\cos x - \sin x\bigr)^2 = 2 .

Solution

Solution of Exercise 14.9.

Expanding both squares and using $\cos^2 x + \sin^2 x = 1$ :

(\cos^2 x + 2\sin x\cos x + \sin^2 x) + (\cos^2 x - 2\sin x\cos x + \sin^2 x) = 1 + 1 = 2 .

Exercise 14.10 ★★

A wheel of radius $30$ cm rolls without slipping. Through what angle (in radians, then in degrees) does it turn when the bike advances by $1.5$ m? How far does the bike advance during one full turn of the wheel?

Solution

Solution of Exercise 14.10.

Rolling without slipping means the arc length equals the distance travelled: $r\theta = 150$ cm gives $\theta = \frac{150}{30} = 5$ rad $= 5 \times \frac{180}{\pi} \approx 286.5^\circ$ . One full turn advances the bike by the circumference $2\pi \times 30 = 60\pi \approx 188.5$ cm, about $1.88$ m.

Exercise 14.11 ★★★

Solve on $\intoc{-\pi}{\pi}$ the inequality $\cos t \leq \frac12$ . (Sketch the unit circle, mark the region where the abscissa is at most $\frac12$ , and read off the interval of values of $t$ .)

Solution

Solution of Exercise 14.11.

On the unit circle, the points of abscissa exactly $\frac12$ are $M\!\left(\pm\frac\pi3\right)$ . The points with abscissa at most $\frac12$ form the arc on the left of the vertical line $x = \frac12$ , travelled as $t$ goes from $\frac\pi3$ up to $\pi$ , or from $-\pi$ up to $-\frac\pi3$ . Hence, on $\intoc{-\pi}{\pi}$ :

S = \intoc{-\pi}{-\frac{\pi}{3}} \cup \intcc{\frac{\pi}{3}}{\pi}.

(The endpoints $\pm\frac\pi3$ are included since the inequality is wide.)

14.6 Problem: The Ferris wheel and the tide

Problem 14.1

Weekend problem — radians as rolled distance, and the unit circle as the master clock of every periodic phenomenon

A Ferris wheel carries you around a circle; the tide carries the harbor’s water up and down the same mathematical circle. Every periodic phenomenon — wheels, tides, sound, seasons — reads its schedule off the unit circle of this chapter (Definition 14.5), and the equations of Method 14.12 tell the exact times. This problem converts, winds, models and solves — and settles on the way why mathematicians measure angles in radians.

Part I — Radians, the rolled distance.

Convert to radians: $30^\circ$ , $135^\circ$ , $300^\circ$ ; and to degrees: $\frac{3\pi}{4}$ , $\frac{5\pi}{6}$ (Method 14.2).
The whole point of radians: an arc of angle $t$ (in radians) on a circle of radius $r$ has length exactly $r\,t$ . What arc does an angle of $2$ radians cut on a circle of radius $3$ m? (No $\pi$ anywhere — that is the point.)
A wheel of radius $30$ cm rolls $1$ km without slipping. Through how many radians has it turned, and how many full turns is that?
Place on the unit circle and evaluate exactly: $\cos\frac{2\pi}{3}$ , $\sin\left(-\frac{\pi}{4}\right)$ , and $\cos\frac{19\pi}{6}$ (reduce modulo $2\pi$ first; Proposition 14.9).
An angle $t \in \intoo{\frac\pi2}{\pi}$ satisfies $\sin t = \frac35$ . Using $\cos^2 t + \sin^2 t = 1$ and the quadrant, find $\cos t$ and $\tan t$ .

Part II — The Ferris wheel. A giant wheel has radius $60$ m, its hub $65$ m above the ground, and turns once every $30$ minutes. You board at the lowest point at time $t = 0$ (in minutes), so your height obeys

h(t) = 65 - 60 \cos\!\left(\frac{\pi t}{15}\right).

Justify the formula: check the angle turned after $t$ minutes, the height at $t = 0$ , and explain the minus sign.
Compute your height at $t = 7.5$ , $t = 15$ and $t = 20$ minutes (exact values).
At what time are you first at $95$ m? (Solve $h(t) = 95$ with Method 14.12.)
During which time interval of the ride are you at least $95$ m high — and for how long in total? (Compare Exercise 14.11.)
Where does the height change fastest? Compare the climb during the first minute ( $h(1) - h(0)$ ) with the climb between minutes $7$ and $8$ , and formulate the observation (the tool that makes it exact — the derivative of the sinusoid — arrives in grade 12).
Engineer’s turn: design the height formula of a wheel with maximal height $40$ m, minimal height $4$ m and period $12$ minutes, boarding at the bottom at $t = 0$ .

Part III — The tide and the circle’s symmetries. In a harbor, the water depth in meters, $t$ hours after midnight, is

d(t) = 7 + 3 \sin\!\left(\frac{\pi t}{6}\right).

Give the maximal and minimal depths, the times at which they first occur, and the period of the tide.
A cargo ship needs $8.5$ m of water. Solve $d(t) \geq 8.5$ on $\intco{0}{12}$ : during which window can it enter, and how long is the window? When does the next window open?
Re-derive two associated-angle identities directly from circle symmetries — $\sin(\pi - t) = \sin t$ (mirror across the vertical axis) and $\cos(\pi + t) = -\cos t$ (half-turn) — and state the complementary identity $\cos\left(\frac\pi2 - t\right) = \sin t$ .
Parity (Problem 11.1 vocabulary): classify $\sin$ , $\cos$ , and $t \mapsto \sin t + t^3$ as even or odd, with one-line proofs from the circle.
Solve on $\intco{0}{2\pi}$ : $2\sin^2 t - \sin t - 1 = 0$ . (A quadratic in disguise: factor it as in Problem 10.1, then read the circle.)

Part IV — The mathematician’s angle.

How many degrees is $1$ radian? Then a calculator trap: which is larger, $\sin(1)$ (radian mode) or $\sin(1^\circ)$ — and by roughly what factor? Moral for calculator settings?
Solve $\cos t = \sin t$ on $\intco{0}{2\pi}$ , using the circle (where are abscissa and ordinate equal?).
For small angles, the arc and its sine almost coincide: compare $\sin(0.1)$ with $0.1$ (five decimals), and give the relative error. This “small-angle approximation” steers ships and pendulums; the theorem behind it ( $\frac{\sin t}{t} \to 1$ ) is a star of the grade-12 limits chapter.
Finale — the unit circle as master clock, one sentence per item: radians turn angles into honest numbers (arc $=$ radius $\times$ angle); cosine and sine are the clock hand’s shadow and height; the circle’s symmetries are the identities; solving $\cos t = c$ reads times off the clock; and every periodic phenomenon — wheel, tide, sound — is this clock wearing a costume.

Solution

Solution of Problem 14.1.

1. $30^\circ = \frac{\pi}{6}$ ; $135^\circ = \frac{3\pi}{4}$ ; $300^\circ = \frac{5\pi}{3}$ . And $\frac{3\pi}{4} = 135^\circ$ ; $\frac{5\pi}{6} = 150^\circ$ .

2. Length $r\,t = 3 \times 2 = 6$ m. Radians make arc length a plain product — degrees would drag a $\frac{\pi}{180}$ into every formula.

3. Radians turned $= \frac{\text{distance}} {\text{radius}} = \frac{1000}{0.30} \approx 3\,333$ rad; divided by $2\pi$ : about $530$ full turns.

4. $\cos\frac{2\pi}{3} = -\frac12$ ; $\sin\left(-\frac\pi4\right) = -\frac{\sqrt2}{2}$ ; $\frac{19\pi}{6} - 2\pi = \frac{7\pi}{6}$ , so $\cos\frac{19\pi}{6} = \cos\frac{7\pi}{6} = -\frac{\sqrt3}{2}$ .

5. $\cos^2 t = 1 - \frac{9}{25} = \frac{16}{25}$ ; in the second quadrant the cosine is negative: $\cos t = -\frac45$ , and $\tan t = \frac{3/5}{-4/5} = -\frac34$ .

6. In $t$ minutes the wheel turns $\frac{t}{30} \times 2\pi = \frac{\pi t}{15}$ . At $t = 0$ : $h = 65 - 60 = 5$ m — the boarding platform at the bottom (hub height minus radius). The minus sign puts you below the hub when the turned angle is $0$ .

7. $h(7.5) = 65 - 60\cos\frac{\pi}{2} = 65$ m (hub height); $h(15) = 65 + 60 = 125$ m (the top); $h(20) = 65 - 60\cos\frac{4\pi}{3} = 65 + 30 = 95$ m.

8. $65 - 60\cos\frac{\pi t}{15} = 95$ gives $\cos\frac{\pi t}{15} = -\frac12$ : first solution $\frac{\pi t}{15} = \frac{2\pi}{3}$ , i.e. $t = 10$ minutes.

9. $\cos\frac{\pi t}{15} \leq -\frac12$ for $\frac{\pi t}{15} \in \intcc{\frac{2\pi}{3}}{\frac{4\pi}{3}}$ , i.e. $t \in \intcc{10}{20}$ : ten minutes of the ride above $95$ m, symmetric about the summit at $t = 15$ .

10. First minute:

h(1) - h(0) = 60\left(1 - \cos\frac{\pi}{15}\right) \approx 1.3 \text{ m}.

Between minutes $7$ and $8$ :

60\left(\cos\frac{7\pi}{15} - \cos\frac{8\pi}{15}\right) \approx 12.5 \text{ m}

— nearly ten times more. The height changes fastest when passing hub level and stalls near bottom and top: the sinusoid is steep at its middle, flat at its extremes.

11. Radius $= \frac{40 - 4}{2} = 18$ m, hub at $4 + 18 = 22$ m, period $12$ min: $h(t) = 22 - 18\cos\left(\frac{\pi t}{6}\right)$ .

12. Maximum $7 + 3 = 10$ m when $\sin\frac{\pi t}{6} = 1$ : first at $t = 3$ (3 a.m.); minimum $4$ m first at $t = 9$ ; period $\frac{2\pi}{\pi/6} = 12$ hours.

13. $\sin\frac{\pi t}{6} \geq \frac12$ for $\frac{\pi t}{6} \in \intcc{\frac\pi6}{\frac{5\pi}{6}}$ : $t \in \intcc{1}{5}$ — a four-hour window from 1 a.m. to 5 a.m., reopening one period later, $t \in \intcc{13}{17}$ (1 p.m. to 5 p.m.).

14. Mirror across the vertical axis sends the point at angle $t$ to the point at angle $\pi - t$ , preserving the ordinate: $\sin(\pi - t) = \sin t$ . The half-turn about the center sends $t$ to $\pi + t$ , negating both coordinates: $\cos(\pi + t) = -\cos t$ . And the mirror across the diagonal swaps abscissa and ordinate: $\cos\left(\frac\pi2 - t\right) = \sin t$ — the “co” of complementary angles.

15. $\sin(-t) = -\sin t$ : odd (half-turn symmetry of its graph); $\cos(-t) = \cos t$ : even (mirror). And $\sin t + t^3$ is a sum of two odd functions: odd.

16. Factor: $(2\sin t + 1)(\sin t - 1) = 0$ : $\sin t = 1$ gives $t = \frac\pi2$ ; $\sin t = -\frac12$ gives $t = \frac{7\pi}{6}$ and $t = \frac{11\pi}{6}$ . Three solutions on the circle.

17. $1$ rad $= \frac{180}{\pi} \approx 57.3^\circ$ . So $\sin(1) \approx 0.841$ while $\sin(1^\circ) \approx 0.0175$ : a factor of nearly $50$ . Moral: check the mode light before trusting any trigonometric display.

18. Abscissa equals ordinate on the diagonal: $t = \frac\pi4$ and $t = \frac{5\pi}{4}$ .

19. $\sin(0.1) = 0.09983\ldots$ against $0.1$ : relative error about $0.17\,\%$ . For small angles the chord hugs the arc — and the grade-12 limit $\frac{\sin t}{t} \to 1$ is exactly this observation made into a theorem.

20. Radians: angle $=$ arc on the unit circle, so angles compute like lengths. Cosine and sine: the coordinates of the clock hand — shadow on the ground, height on the wall. Symmetries of the circle: the whole catalogue of identities, read off mirrors and half-turns. Equations $\cos t = c$ : the clock’s timetable, with the circle showing every solution at once. Costumes: the wheel wore $65 - 60\cos$ , the tide wore $7 + 3\sin$ — same clock, same mathematics.