International Baccalaureate · DP 2: Applications and Interpretation SL

Trigonometric Functions

⌚ ~106 min ▢ 11 blocks ✓ 40 exercises ➣ Prerequisites : Equations, Non-Right-Angled Triangle Trigonometry, Functions, Modelling

I Periodic Behaviour

I.1 Periodic Functions

A function $f$ is periodic if it repeats itself in regular intervals.

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Definition — Periodic Function

A function $f$ is periodic if there exists a constant $P>0$ such that$$f(x+P) = f(x), \forall x$$

For a periodic function that has a smallest such constant, the period is this smallest value of $P$: the length of one full repetition, or cycle.
For a periodic function that has a maximum value and a minimum value, the principal axis is the horizontal line about which the curve oscillates, given by $$y = \frac{\text{maximum value} + \text{minimum value}}{2}$$
For such a function, the amplitude is the distance from the principal axis to a maximum or minimum point, given by $$\frac{\text{maximum value} - \text{minimum value}}{2}$$

Among the many types of periodic functions, some exhibit a smooth, wave-like pattern. These sinusoidal functions are modelled with the sine and cosine trigonometric functions.

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Skills to practice

Recognising Periodic Behaviour

II The Unit Circle

II.1 Cosine and Sine of Any Angle

Recappreviously studied — click to reread

Definition — Unit circle

The unit circle is the circle with a radius of 1 centred at the origin: it is the circle used to measure angles, travelled anticlockwise.

Proposition — Relationship between Angle and Coordinates

For an angle $\theta$, measured anticlockwise from the positive $x$-axis, $M(x,y)$ is the corresponding point on the unit circle.

The $x$-coordinate is the cosine of the angle: $\cos \theta = x$
The $y$-coordinate is the sine of the angle: $\sin \theta = y$

For an acute angle, these equalities are proved with the right-angled triangle formed by $O$, $M$ and the foot of the perpendicular from $M$ to the $x$-axis. For any other angle, these coordinates define $\cos\theta$ and $\sin\theta$.

Proof

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

Skills to practice

Cosine and Sine of Any Angle

III The Sine and Cosine Functions in Degrees

III.1 The Curves of $y \equal \sin x$ and $y \equal \cos x$

Angles are often measured in degrees: on a compass, in a triangle, on a turning wheel. The unit circle gives the cosine and the sine of every angle, whatever its unit: the point of the circle for the angle $x$ has coordinates $(\cos x, \sin x)$. Let the angle $x$ run through the degrees, from $0^\circ$ to $360^\circ$ and beyond, and follow each coordinate on its own: this draws the curves of $y = \sin x$ and $y = \cos x$ above a horizontal axis graduated in degrees.

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Proposition — The Sine and Cosine Functions in Degrees

For an angle $x$ measured in degrees:

$\sin x$ and $\cos x$ are defined for every angle $x$, and $-1 \le \sin x \le 1$, $-1 \le \cos x \le 1$: the range of each function is the interval from $-1$ to $1$;
the two functions are periodic with period $360^\circ$: for every angle $x$ and every integer $k$, $$\sin(x + 360^\circ k) = \sin x \qquad \text{and} \qquad \cos(x + 360^\circ k) = \cos x;$$
for $0^\circ \le x \le 360^\circ$, $y = \sin x$ starts at $0$, reaches its maximum $1$ at $90^\circ$, returns to $0$ at $180^\circ$, reaches its minimum $-1$ at $270^\circ$ and returns to $0$ at $360^\circ$; $y = \cos x$ starts at its maximum $1$ at $0^\circ$, is $0$ at $90^\circ$, reaches its minimum $-1$ at $180^\circ$, is $0$ at $270^\circ$ and returns to $1$ at $360^\circ$.

Proof

The point $M$ of the unit circle for the angle $x$ has coordinates $(\cos x, \sin x)$ (Relationship between Angle and Coordinates).

$M$ lies on the circle of centre $O$ and radius $1$, so each of its coordinates lies between $-1$ and $1$. Every value $c$ from $-1$ to $1$ is reached: the horizontal line $y = c$ meets the circle, and the angle of a meeting point has sine $c$; in the same way the vertical line $x = c$ gives an angle with cosine $c$.
Adding $360^\circ$ to the angle adds a full turn, which brings $M$ back to the same point: the angles $x$ and $x + 360^\circ$ have the same coordinates. Repeating this $k$ times (turning the other way when $k$ is negative) gives $x + 360^\circ k$. No smaller positive angle $P$ works for every $x$: the only angles of sine $1$ are $90^\circ + 360^\circ k$ (the point $(0, 1)$), so $\sin(90^\circ + P) = 1$ forces $P$ to be a multiple of $360^\circ$; in the same way $\cos(0^\circ + P) = 1$ forces it for the cosine.
The points for $0^\circ$, $90^\circ$, $180^\circ$, $270^\circ$ and $360^\circ$ are $(1, 0)$, $(0, 1)$, $(-1, 0)$, $(0, -1)$ and $(1, 0)$: their $y$-coordinates give the values of the sine, their $x$-coordinates those of the cosine.

Example

Complete the table of values of $y = \sin x$, giving each value that is not exact to $2$ decimal places, then draw the graph of $y = \sin x$ for $0^\circ \le x \le 360^\circ$.

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$
$\sin x$

$x$	$210^\circ$	$240^\circ$	$270^\circ$	$300^\circ$	$330^\circ$	$360^\circ$
$\sin x$

Answer

From $0^\circ$ to $90^\circ$ the values are the exact values: $\sin 0^\circ = 0$, $\sin 30^\circ = \dfrac{1}{2}$, $\sin 60^\circ = \dfrac{\sqrt{3}}{2} \approx 0.87$, $\sin 90^\circ = 1$. The other angles are read on the unit circle, where the points for $30^\circ$, $150^\circ$, $210^\circ$ and $330^\circ$ are placed symmetrically:

the point for $180^\circ - x$ is the reflection of the point for $x$ in the vertical axis: same $y$-coordinate, so $\sin 150^\circ = \sin 30^\circ$ and $\sin 120^\circ = \sin 60^\circ$;
the point for $180^\circ + x$ is the image of the point for $x$ by the half-turn about $O$: opposite $y$-coordinate, so $\sin 210^\circ = -\sin 30^\circ$ and $\sin 240^\circ = -\sin 60^\circ$;
the point for $360^\circ - x$ is the reflection of the point for $x$ in the horizontal axis: opposite $y$-coordinate, so $\sin 330^\circ = -\sin 30^\circ$ and $\sin 300^\circ = -\sin 60^\circ$.

$x$	$0^\circ$	$30^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$150^\circ$	$180^\circ$
$\sin x$	$0$	$0.5$	$0.87$	$1$	$0.87$	$0.5$	$0$

$x$	$210^\circ$	$240^\circ$	$270^\circ$	$300^\circ$	$330^\circ$	$360^\circ$
$\sin x$	$-0.5$	$-0.87$	$-1$	$-0.87$	$-0.5$	$0$

A calculator set to degree mode gives the same values. Plotting the points and joining them with a smooth curve:

Example

Using the period, find the exact values of $\cos 420^\circ$, $\cos(-60^\circ)$ and $\sin 750^\circ$.
Find every angle $x$ with $0^\circ \le x \le 720^\circ$ such that $\cos x = \dfrac{1}{2}$.

Answer

Each angle is brought back between $0^\circ$ and $360^\circ$ by whole turns, which do not change the cosine or the sine: $$\begin{aligned}[t] \cos 420^\circ &= \cos(60^\circ + 360^\circ) &&\text{(}420 = 60 + 360\text{)}\\ &= \cos 60^\circ &&\text{(period } 360^\circ\text{)}\\ &= \frac{1}{2}, \end{aligned} \qquad \begin{aligned}[t] \sin 750^\circ &= \sin(30^\circ + 2 \times 360^\circ) &&\text{(}750 = 30 + 720\text{)}\\ &= \sin 30^\circ &&\text{(period } 360^\circ\text{)}\\ &= \frac{1}{2}. \end{aligned}$$ For $\cos(-60^\circ)$, adding one turn gives $\cos(-60^\circ) = \cos 300^\circ$. The point for $300^\circ = 360^\circ - 60^\circ$ is the reflection of the point for $60^\circ$ in the horizontal axis: same $x$-coordinate, so $\cos(-60^\circ) = \cos 60^\circ = \dfrac{1}{2}$.
Between $0^\circ$ and $360^\circ$, the vertical line $x = \dfrac{1}{2}$ meets the unit circle at two points: the point for $60^\circ$ (as $\cos 60^\circ = \dfrac{1}{2}$) and its reflection in the horizontal axis, the point for $360^\circ - 60^\circ = 300^\circ$. Adding one turn to each gives the solutions between $360^\circ$ and $720^\circ$: $60^\circ + 360^\circ = 420^\circ$ and $300^\circ + 360^\circ = 660^\circ$. The solutions are $$x = 60^\circ, \quad x = 300^\circ, \quad x = 420^\circ, \quad x = 660^\circ.$$ On the graph, they are the four points where the curve of $y = \cos x$ meets the line $y = \dfrac{1}{2}$:

Skills to practice

Reading the Curves in Degrees
Using the Period in Degrees
Finding Exact Values beyond $180^\circ$

IV General Sine and Cosine Functions

IV.1 Amplitude, Period and Principal Axis

Proposition — General Sine and Cosine Functions in Degrees

Let $a$, $b$ and $d$ be real numbers with $a \neq 0$ and $b \neq 0$. When $x$ is measured in degrees, the functions $$y = a\sin(bx) + d \qquad \text{and} \qquad y = a\cos(bx) + d$$ have

amplitude $|a|$;
period $\dfrac{360^\circ}{|b|}$;
principal axis the line $y = d$;
maximum value $d + |a|$ and minimum value $d - |a|$.

Proof

We write the proof for $f(x) = a\sin(bx) + d$; replacing $\sin$ by $\cos$ changes nothing below.

Values. As $x$ runs through all the angles, so does $bx$ (as $b \neq 0$), and $\sin(bx)$ takes every value from $-1$ to $1$, and only those. Multiplying by $a$ gives every value from $-|a|$ to $|a|$ (in the reverse order when $a < 0$); adding $d$ gives every value from $d - |a|$ to $d + |a|$. So the maximum value is $d + |a|$ and the minimum value $d - |a|$. By the definitions of the principal axis and of the amplitude, the principal axis is $y = \dfrac{(d + |a|) + (d - |a|)}{2} = d$ and the amplitude is $\dfrac{(d + |a|) - (d - |a|)}{2} = |a|$.
Period. Let $P = \dfrac{360^\circ}{|b|}$. Then $bP = \pm 360^\circ$, and since the sine has period $360^\circ$, $$\begin{aligned}[t] f(x + P) &= a\sin(bx + bP) + d \\ &= a\sin(bx \pm 360^\circ) + d \\ &= a\sin(bx) + d && \text{(period } 360^\circ \text{ of the sine)}\\ &= f(x). \end{aligned}$$ No smaller positive number works: if $f(x + Q) = f(x)$ for every $x$, with $Q > 0$, then $\sin(bx + bQ) = \sin(bx)$ for every $x$ (divide by $a \neq 0$), that is $\sin(u + |b|Q) = \sin u$ for every angle $u$; so $|b|Q$ is at least the period $360^\circ$ of the sine, and $Q \ge \dfrac{360^\circ}{|b|} = P$.

Example

Consider the function $y = 3\sin(2x) + 1$, with $x$ in degrees.

State its amplitude, its period and its principal axis.
State its maximum and minimum values, and the values of $x$ with $0^\circ \le x \le 360^\circ$ at which they are reached.
Sketch its graph for $0^\circ \le x \le 360^\circ$.

Answer

Here $a = 3$, $b = 2$ and $d = 1$.

Amplitude $|a| = 3$; period $\dfrac{360^\circ}{|b|} = \dfrac{360^\circ}{2} = 180^\circ$; principal axis $y = 1$.
Maximum value $d + |a| = 1 + 3 = 4$, minimum value $d - |a| = 1 - 3 = -2$. At $x = 0^\circ$ the curve is on its principal axis, rising, as $y = 3\sin 0^\circ + 1 = 1$; it reaches its maximum a quarter of a period later and its minimum three quarters of a period later: $$\begin{aligned}[t] x &= \tfrac{1}{4} \times 180^\circ = 45^\circ && \text{(first maximum)}\\ x &= \tfrac{3}{4} \times 180^\circ = 135^\circ && \text{(first minimum)} \end{aligned}$$ Adding the period $180^\circ$ gives the others: the maximum is reached at $x = 45^\circ$ and $x = 225^\circ$, the minimum at $x = 135^\circ$ and $x = 315^\circ$; one more period goes beyond $360^\circ$. Check: $3\sin 90^\circ + 1 = 4$ and $3\sin 270^\circ + 1 = -2$.
The curve starts on its principal axis at $(0^\circ, 1)$, rising, and repeats every $180^\circ$:

Example

The curve below has equation $y = a\cos(bx) + d$, with $a > 0$, $b > 0$ and $x$ in degrees. It has a maximum point at $(0^\circ, 5)$, and its next minimum point is $(60^\circ, -1)$. Find $a$, $b$ and $d$.

Answer

The maximum value is $5$ and the minimum value is $-1$.

Principal axis: $d = \dfrac{5 + (-1)}{2} = 2$.
Amplitude: $|a| = \dfrac{5 - (-1)}{2} = 3$, and $a > 0$, so $a = 3$.
From a maximum to the next minimum is half a period: the period is $2 \times 60^\circ = 120^\circ$. As $b > 0$, $\dfrac{360^\circ}{b} = 120^\circ$, so $b = \dfrac{360}{120} = 3$.

The equation is $y = 3\cos(3x) + 2$. Check: at $x = 0^\circ$, $y = 3\cos 0^\circ + 2 = 5$; at $x = 60^\circ$, $y = 3\cos 180^\circ + 2 = -3 + 2 = -1$.

Skills to practice

Reading Period, Principal Axis and Amplitude
Finding a Sinusoid from its Graph in Degrees

V Modelling Periodic Behaviour

V.1 Solving Equations with Technology

Recappreviously studied — click to reread

Method — Solving an Equation Graphically with Technology

The solutions of an equation $f(x) = g(x)$ are the $x$-coordinates of the points where the graphs of $f$ and $g$ meet; the solutions of an equation $f(x) = 0$ are the $x$-intercepts of the graph of $f$.

Enter each side as a function: $Y_1 = f(x)$ and $Y_2 = g(x)$. For an equation $f(x) = 0$, enter $f(x)$ alone.
Choose a window that shows every point of intersection, or every $x$-intercept. A table of values of $f(x) - g(x)$ helps: a change of sign between two consecutive values of $x$ signals a solution between them, provided the curves have no break between these two values: this holds for polynomials, and for rational functions away from the values that make a denominator zero. Where the two curves touch without crossing, $f(x) - g(x)$ does not change sign: such a solution is seen only on the graph.
In the analysis tools of the graph menu, choose intersection (for $f(x) = 0$: zero), select the curves and give a first guess, or bounds, close to one point. Read the $x$-coordinate $X$ of the point displayed: the $y$-coordinate $Y$ is the common value of the two sides, not a solution.
Repeat for each point, then report the solutions: exactly when the screen shows an exact value that substitution confirms, otherwise to 3 significant figures.

The mock-up shows the equation $x^2 - 2 = \dfrac{1}{x}$: the two curves meet three times, and the point read is the one on the right.

Skills to practice

Solving Equations with Technology

V.2 Sinusoidal Models in Degrees

Common real-world examples of sinusoidal models include phenomena that oscillate (fluctuate periodically):

$D(t)$ is the depth of water at a shore $t$ hours after midnight
$T(n)$ is the temperature of a city $n$ days after January 1st
$H(t)$ is the vertical height above ground of a person $t$ seconds after entering a Ferris wheel

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Method — Modelling Periodic Data in Degrees

A quantity that repeats itself every $P$ units of time in a smooth wave, rising and falling symmetrically about a middle value, can be modelled by $y = a\sin(bt) + d$ or $y = a\cos(bt) + d$, where the angle $bt$ is measured in degrees.

Principal axis: $d = \dfrac{\text{maximum} + \text{minimum}}{2}$.
Amplitude: $|a| = \dfrac{\text{maximum} - \text{minimum}}{2}$.
Period: one period is one full turn of $360^\circ$, so $b = \dfrac{360}{P}$, in degrees per unit of time ($b > 0$).
Form and sign, read on the start of the data, at $t = 0$:
- on the principal axis and rising: $y = a\sin(bt) + d$ with $a > 0$;
- on the principal axis and falling: $y = a\sin(bt) + d$ with $a < 0$;
- at a maximum: $y = a\cos(bt) + d$ with $a > 0$;
- at a minimum: $y = a\cos(bt) + d$ with $a < 0$.
The time origin is chosen so that the data start at one of these four positions; data that start anywhere else need a horizontal translation of the curve.
Check the model on a data point, and set the calculator to degree mode before using it.

Example

In a city of the southern hemisphere, the mean monthly temperature is highest in January, $26^\circ$C, and lowest in July, $10^\circ$C, and the pattern repeats every year. Let $t$ be the time in months, with $t = 0$ in January and $t = 6$ in July.

Find a model $T(t) = a\cos(bt) + d$ for the mean monthly temperature, the angle $bt$ being in degrees.
Use the model to estimate the mean temperature in April ($t = 3$) and in May ($t = 4$).

Answer

Following the Method:
- principal axis: $d = \dfrac{26 + 10}{2} = 18$;
- amplitude: $|a| = \dfrac{26 - 10}{2} = 8$;
- the period is $12$ months, so $b = \dfrac{360}{12} = 30$ (degrees per month);
- at $t = 0$ the temperature is at its maximum: the form $a\cos(bt) + d$ with $a > 0$, so $a = 8$.
The model is $T(t) = 8\cos(30t) + 18$. Check in July: $T(6) = 8\cos 180^\circ + 18 = -8 + 18 = 10$, the minimum of the data.
$$\begin{aligned}[t] T(3) &= 8\cos 90^\circ + 18 = 8 \times 0 + 18 = 18,\\ T(4) &= 8\cos 120^\circ + 18 = 8 \times \left(-\frac{1}{2}\right) + 18 = 14. \end{aligned}$$ The model estimates about $18^\circ$C in April and $14^\circ$C in May.

Example

A Ferris wheel has a diameter of $40$ m and its axle is $22$ m above the ground. It turns at a constant rate, once every $12$ minutes. A passenger boards at the lowest point of the wheel at $t = 0$, $t$ in minutes.

Find a model $h(t) = a\cos(bt) + d$ for the height of the passenger above the ground, in metres, the angle $bt$ being in degrees.
Using a graphing calculator, find when the passenger first reaches $35$ m above the ground, and for how long, during each turn, the passenger is higher than $35$ m.

Answer

The lowest point is $22 - 20 = 2$ m above the ground and the highest $22 + 20 = 42$ m.
- principal axis: $d = \dfrac{42 + 2}{2} = 22$, the height of the axle;
- amplitude: $|a| = \dfrac{42 - 2}{2} = 20$, the radius;
- period $12$ minutes: $b = \dfrac{360}{12} = 30$ (degrees per minute);
- at $t = 0$ the passenger is at the lowest point, a minimum: the form $a\cos(bt) + d$ with $a < 0$, so $a = -20$.
The model is $h(t) = -20\cos(30t) + 22$. Check at the top, half a turn later: $h(6) = -20\cos 180^\circ + 22 = 20 + 22 = 42$.
In degree mode, enter $Y_1 = -20\cos(30X) + 22$ and $Y_2 = 35$, with the window $0 \le X \le 12$, $0 \le Y \le 45$: one turn. The line meets the curve twice (Method Solving an Equation Graphically with Technology): The intersection tool gives $X \approx 4.3513867$ and then $X \approx 7.6486133$. The passenger first reaches $35$ m after about $\boldsymbol{4.35}$ minutes (about $4$ min $21$ s), and is higher than $35$ m between the two times: $$7.6486133 - 4.3513867 = 3.2972266 \approx \boldsymbol{3.30} \text{ minutes},$$ about $3$ min $18$ s in each turn. The two times are symmetric about $t = 6$, the top of the wheel: $4.3513867 + 7.6486133 = 12$.

Skills to practice

Building a Sinusoidal Model in Degrees
Solving Equations in Degree Mode with Technology

VI Going further

Skills to practice

Solving Problems with Sinusoidal Models
Quiz

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)
\(\sin x\)

\(x\)	\(210^\circ\)	\(240^\circ\)	\(270^\circ\)	\(300^\circ\)	\(330^\circ\)	\(360^\circ\)
\(\sin x\)

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)
\(\sin x\)	\(0\)	\(0.5\)	\(0.87\)	\(1\)	\(0.87\)	\(0.5\)	\(0\)

\(x\)	\(210^\circ\)	\(240^\circ\)	\(270^\circ\)	\(300^\circ\)	\(330^\circ\)	\(360^\circ\)
\(\sin x\)	\(-0.5\)	\(-0.87\)	\(-1\)	\(-0.87\)	\(-0.5\)	\(0\)