---
title: "Trigonometry: The Unit Circle"
book: "High School Mathematics"
subject: math
language: en
chapter: 14
exercises: 11
source: https://one-course.com/books/math/2/en/chapter/14-trigonometry-the-unit-circle
---

# Chapter 14 — Trigonometry: The Unit Circle

Trigonometry begins in right triangles, but its true home is the *unit circle*, where [cosine](#def-g11-trigo-cossin) and [sine](#def-g11-trigo-cossin) become [functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) of an arbitrary [real number](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets). This chapter installs the [radian](#def-g11-trigo-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](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-function) — variations, [derivatives](https://one-course.com/books/math/2/en/chapter/12-differentiation-a-first-course#def-g11-deriv-derivative) — is carried out in [Chapter 24](https://one-course.com/books/math/2/en/chapter/24-trigonometric-functions#ch-g12-trigo).

## 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](#def-g11-trigo-radian) are proportional: multiply by $\frac{\pi}{180}$ to go from degrees to [radians](#def-g11-trigo-radian), 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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $t$, 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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.](https://one-course.com/images/onecourse/chapters/math-2/g11-trigo/fig-ed4ec1d99063.svg)

*The unit circle: the [real number](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $t$, wound counterclockwise from $I$, lands at the point $M(t)$ whose [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) *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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) are between $-1$ and $1$, and they satisfy the circle’s [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](#def-g11-trigo-winding)). ∎

**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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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](#def-g11-trigo-cossin) read $\frac{\sqrt0}{2}, \frac{\sqrt1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2},
\frac{\sqrt4}{2}$, and the [cosines](#def-g11-trigo-cossin) 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) is unchanged, the [ordinate](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) changes sign, the [ordinate](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) is unchanged.

$M(\pi + t)$ is diametrically opposite $M(t)$ (half a turn): both [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system). ∎

![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.](https://one-course.com/images/onecourse/chapters/math-2/g11-trigo/fig-da7bcec307c7.svg)

*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](#prop-g11-trigo-associated) 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\cos t = \cos acost=cosa and sin⁡t=sin⁡a\sin t = \sin asint=sina).**

For [real numbers](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $t$ 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) exactly when they are equal or reflections of each other in the $x$-axis; by [Proposition 14.9](#prop-g11-trigo-associated), 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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\cos t = ccost=c on an interval).**

1. Find *one* angle $a$ with $\cos a = c$ , from the table of known values.
2. Write the general solutions $t = \pm a + 2k\pi$ ( [Theorem 14.11](#thm-g11-trigo-equations) ).
3. Choose the [integers](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets) $k$ that land inside the requested [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-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.](https://one-course.com/images/onecourse/chapters/math-2/g11-trigo/fig-073ce0f8bd3d.svg)

*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](#def-g11-trigo-radian): $30^\circ$, $45^\circ$, $120^\circ$, $270^\circ$. Convert to degrees: $\frac{\pi}{5}$, $\frac{5\pi}{6}$, $\frac{7\pi}{4}$, $\frac{2\pi}{9}$.

**Solution of Exercise 14.1.**

To [radians](#def-g11-trigo-radian): $30^\circ = \frac{\pi}{6}$, $45^\circ = \frac{\pi}{4}$, $120^\circ = \frac{2\pi}{3}$, $270^\circ = \frac{3\pi}{2}$.

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 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](#prop-g11-trigo-associated), 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 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 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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 of Exercise 14.5.**

Using [Proposition 14.9](#prop-g11-trigo-associated):

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

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

**Exercise 14.6 ★★.**

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

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

**Solution 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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval)).

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

$\cos t = -1$: the only point of [abscissa](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $-1$ 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 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $\cos 2t = \cos t$. (Use [Theorem 14.11](#thm-g11-trigo-equations) with $a = t$, and discuss the two families of solutions.)

**Solution of Exercise 14.8.**

By [Theorem 14.11](#thm-g11-trigo-equations), $\cos 2t = \cos t$ 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 of Exercise 14.9.**

[Expanding](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand) 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](#def-g11-trigo-radian), 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 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) is at most $\frac12$, and read off the [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) of values of $t$.)

**Solution of Exercise 14.11.**

On the unit circle, the points of [abscissa](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) exactly $\frac12$ are $M\!\left(\pm\frac\pi3\right)$. The points with [abscissa](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) *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](#def-g11-trigo-cossin)), and the [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of [Method 14.12](#met-g11-trigo-solve) tell the exact times. This problem converts, winds, models and solves — and settles on the way why mathematicians measure angles in [radians](#def-g11-trigo-radian).

**Part I — [Radians](#def-g11-trigo-radian), the rolled distance.**

1. Convert to [radians](#def-g11-trigo-radian) : $30^\circ$ , $135^\circ$ , $300^\circ$ ; and to degrees: $\frac{3\pi}{4}$ , $\frac{5\pi}{6}$ ( [Method 14.2](#met-g11-trigo-convert) ).
2. The whole point of [radians](#def-g11-trigo-radian) : an arc of angle $t$ (in [radians](#def-g11-trigo-radian) ) on a circle of radius $r$ has length exactly $r\,t$ . What arc does an angle of $2$ [radians](#def-g11-trigo-radian) cut on a circle of radius $3$ m? (No $\pi$ anywhere — that is the point.)
3. A wheel of radius $30$ cm rolls $1$ km without slipping. Through how many *[radians](#def-g11-trigo-radian)* has it turned, and how many full turns is that?
4. 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](#prop-g11-trigo-associated) ).
5. 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

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

6. Justify the formula: check the angle turned after $t$ minutes, the height at $t = 0$ , and explain the minus sign.
7. Compute your height at $t = 7.5$ , $t = 15$ and $t = 20$ minutes (exact values).
8. At what time are you *first* at $95$ m? (Solve $h(t) = 95$ with [Method 14.12](#met-g11-trigo-solve) .)
9. During which time [interval](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interval) of the ride are you at least $95$ m high — and for how long in total? (Compare [Exercise 14.11](#exo-g11-trigo-11) .)
10. 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](https://one-course.com/books/math/2/en/chapter/12-differentiation-a-first-course#def-g11-deriv-derivative) of the sinusoid — arrives in grade 12).
11. 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

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

12. Give the maximal and minimal depths, the times at which they first occur, and the period of the tide.
13. 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?
14. 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 $\cos\left(\frac\pi2 - t\right) = \sin t$ .
15. Parity ( [Problem 11.1](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#pb-g11-func-1) vocabulary): classify $\sin$ , $\cos$ , and $t \mapsto \sin t + t^3$ as even or odd, with one-line proofs from the circle.
16. 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](https://one-course.com/books/math/2/en/chapter/10-quadratic-functions-and-equations#pb-g11-quad-1) , then read the circle.)

**Part IV — The mathematician’s angle.**

17. How many degrees is $1$ [radian](#def-g11-trigo-radian) ? Then a calculator trap: which is larger, $\sin(1)$ ( [radian](#def-g11-trigo-radian) mode) or $\sin(1^\circ)$ — and by roughly what factor? Moral for calculator settings?
18. Solve $\cos t = \sin t$ on $\intco{0}{2\pi}$ , using the circle (where are [abscissa](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) and [ordinate](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) equal?).
19. For small angles, the arc and its [sine](#def-g11-trigo-cossin) almost coincide: compare $\sin(0.1)$ with $0.1$ (five decimals), and give the relative error. This “small-angle [approximation](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-approx) ” steers ships and pendulums; the theorem behind it ( $\frac{\sin t}{t} \to 1$ ) is a star of the grade-12 limits chapter.
20. Finale — the unit circle as master clock, one sentence per item: [radians](#def-g11-trigo-radian) turn angles into honest numbers (arc $=$ radius $\times$ angle); [cosine](#def-g11-trigo-cossin) and [sine](#def-g11-trigo-cossin) 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 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](#def-g11-trigo-radian) make arc length a plain product — degrees would drag a $\frac{\pi}{180}$ into every formula.

**3.** [Radians](#def-g11-trigo-radian) 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](#def-g11-trigo-cossin) 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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $7 + 3 = 10$ m when $\sin\frac{\pi t}{6} = 1$: first at $t = 3$ (3 a.m.); [minimum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) $4$ 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system): $\sin(\pi - t) = \sin t$. The half-turn about the center sends $t$ to $\pi + t$, negating both [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system): $\cos(\pi + t) = -\cos t$. And the mirror across the diagonal swaps [abscissa](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) and [ordinate](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system): $\cos\left(\frac\pi2 - t\right) = \sin t$ — the “co” of complementary angles.

**15.** $\sin(-t) = -\sin t$: odd (half-turn symmetry of its [graph](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-graph)); $\cos(-t) = \cos t$: even (mirror). And $\sin t + t^3$ is a sum of two [odd functions](https://one-course.com/books/math/2/en/chapter/11-functions-and-variations#def-g11-func-parity): 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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) equals [ordinate](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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](#def-g11-trigo-radian): angle $=$ arc on the unit circle, so angles compute like lengths. [Cosine](#def-g11-trigo-cossin) and [sine](#def-g11-trigo-cossin): the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $\cos t = c$: 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.