International Baccalaureate · DP 1: Analysis and Approaches HL

Trigonometric Functions

⌚ ~178 min ▢ 20 blocks ✓ 93 exercises ➣ Prerequisites : The Unit Circle and Radian Measure, Transformations of functions

I Periodic Functions

I.1 Periodic Behaviour

Recappreviously studied — click to reread

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$$f(x+P) = f(x), \forall x$$

The period is the length of one full repetition or cycle. It is the smallest positive value $p$ such that $f(x+p) = f(x)$ for all $x$.
The principal axis or mean line is the horizontal line about which the wave oscillates, given by $$y = \frac{\text{maximum value} + \text{minimum value}}{2}$$
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

Identifying Periodic Behaviour from a Graph

II The Sine and Cosine Functions

II.1 Sine and Cosine as Functions

Recappreviously studied — click to reread

Let $M(\cos x, \sin x)$ be the point on the unit circle corresponding to an angle $x$ (in radians).

The angle $x$ on the unit circle corresponds to the input for the sine function.
The $y$-coordinate of point $M$ on the unit circle, $\sin x$, gives the output of the sine function.

Thus, plotting $x \mapsto \sin x$ produces the graph of the sine function.
See for example: \href{https://www.geogebra.org/m/j7w29vj4}{GeoGebra demo}.

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Definition — Sine Function

The sine function, denoted $\sin$, is the function defined on $\mathbb{R}$ which maps every real number $x$ to the $y$-coordinate of the point $M$ of the trigonometric circle associated with $x$: $x\mapsto \sin(x)$.

Example

Complete the following table with the values of the sine function at key angles:

$x$	0	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$	$\dfrac{2 \pi}{3}$	$\dfrac{3 \pi}{4}$	$\dfrac{5 \pi}{6}$	$\pi$
$\sin (x)$

Answer

$x$	0	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$	$\dfrac{2 \pi}{3}$	$\dfrac{3 \pi}{4}$	$\dfrac{5 \pi}{6}$	$\pi$
$\sin (x)$	0	$\dfrac{1}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{\sqrt{3}}{2}$	1	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{1}{2}$	$0$

If we instead project the values of $\cos x$ from the unit circle onto a graph, we obtain the graph of the cosine function $x \mapsto \cos x$.

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Definition — Cosine Function

The cosine function, denoted $\cos$, is the function defined on $\mathbb{R}$ which maps every real number $x$ to the $x$-coordinate of the point $M$ of the trigonometric circle associated with $x$: $x\mapsto\cos(x)$.

Example

Complete the following table with the values of the cosine function at key angles:

$x$	0	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$	$\dfrac{2 \pi}{3}$	$\dfrac{3 \pi}{4}$	$\dfrac{5 \pi}{6}$	$\pi$
$\cos (x)$

Answer

$x$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$	$\dfrac{2 \pi}{3}$	$\dfrac{3 \pi}{4}$	$\dfrac{5 \pi}{6}$	$\pi$
$\cos (x)$	$1$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{\sqrt{2}}{2}$	$\dfrac{1}{2}$	$0$	$-\dfrac{1}{2}$	$-\dfrac{\sqrt{2}}{2} $	$-\dfrac{\sqrt{3}}{2}$	$-1$

Proposition — Domain and Range of the Sine and Cosine Functions

The sine and cosine functions are defined on $\mathbb{R}$. For every real $x$, $-1 \le \sin x \le 1$ and $-1 \le \cos x \le 1$, and, the point associated with $x$ running over the whole unit circle, its $x$-coordinate and its $y$-coordinate take every value between $-1$ and $1$: the range of each function is $[-1, 1]$.

Proposition — Properties of Sine and Cosine

Periodicity: Both functions are periodic with a period of $2\pi$. $\sin(x+2\pi) = \sin(x)$ and $\cos(x+2\pi) = \cos(x)$.
Parity: Cosine is an even function ($\cos(-x)=\cos(x)$). Sine is an odd function ($\sin(-x)=-\sin(x)$).

Skills to practice

Completing Tables and Reading Graphs
Plotting the Sine and Cosine Curves
Reading Solutions and Periods on the Curves

II.2 Variations and Symmetries of the Curves

Recappreviously studied — click to reread

Periodicity says that the curves of cosine and sine repeat a pattern of length $2\pi$. The trigonometric circle tells more about this pattern: where it rises, where it falls, and which symmetries it has.

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Proposition — Variations of the Cosine and Sine Functions

On $\left[-\pi, \pi\right]$:

the cosine function is increasing on $\left[-\pi, 0\right]$ and decreasing on $\left[0, \pi\right]$;
the sine function is increasing on $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, and decreasing on $\left[-\pi, -\dfrac{\pi}{2}\right]$ and on $\left[\dfrac{\pi}{2}, \pi\right]$.

By periodicity, these tables repeat on every interval $\left[-\pi + 2k\pi, \pi + 2k\pi\right]$, $k$ an integer.

Proof

Read on the trigonometric circle, where the image $M(x)$ has coordinates $(\cos x, \sin x)$.

When $x$ goes from $0$ to $\pi$, $M(x)$ travels the upper half-circle from $(1, 0)$ to $(-1, 0)$: its $x$-coordinate $\cos x$ goes down from $1$ to $-1$. When $x$ goes from $-\pi$ to $0$, $M(x)$ travels the lower half-circle from $(-1, 0)$ to $(1, 0)$: $\cos x$ goes up from $-1$ to $1$.
When $x$ goes from $-\dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$, $M(x)$ travels the right half-circle from $(0, -1)$ to $(0, 1)$: its $y$-coordinate $\sin x$ goes up from $-1$ to $1$. From $\dfrac{\pi}{2}$ to $\pi$, $M(x)$ goes from $(0, 1)$ to $(-1, 0)$ and $\sin x$ goes down from $1$ to $0$; from $-\pi$ to $-\dfrac{\pi}{2}$, it goes from $(-1, 0)$ to $(0, -1)$ and $\sin x$ goes down from $0$ to $-1$.

Proposition — Symmetries of the Cosine and Sine Curves

For every real number $x$, the associated angles give:

$\sin(\pi - x) = \sin x$: the sine curve is symmetric about the vertical line $x = \dfrac{\pi}{2}$;
$\sin(\pi + x) = -\sin x$ and $\sin(\pi - x) = \sin x$: the sine curve is symmetric about the point $(\pi, 0)$;
$\cos(\pi - x) = -\cos x$: the cosine curve is symmetric about the point $\left(\dfrac{\pi}{2}, 0\right)$;
$\cos\left(x - \dfrac{\pi}{2}\right) = \sin x$: the sine curve is the cosine curve translated by $\dfrac{\pi}{2}$ to the right.

Proof

The relations come from the trigonometric circle: the images of $x$ and $\pi - x$ are symmetric about the vertical axis (same $y$-coordinate, opposite $x$-coordinates), so $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$; the images of $x$ and $\pi + x$ are symmetric about the centre (half-turn), so $\sin(\pi + x) = -\sin x$; the images of $x$ and $\dfrac{\pi}{2} - x$ are symmetric about the line $y = x$, which swaps the two coordinates, so $\cos\left(\dfrac{\pi}{2} - x\right) = \sin x$. Two points $(a, b)$ and $(a', b')$ are symmetric about the vertical line $x = m$ when $b' = b$ and $\dfrac{a + a'}{2} = m$, and about the point $(m, 0)$ when $b' = -b$ and $\dfrac{a + a'}{2} = m$.

The points $(x, \sin x)$ and $(\pi - x, \sin(\pi - x)) = (\pi - x, \sin x)$ of the sine curve have the same $y$-coordinate, and $\dfrac{x + (\pi - x)}{2} = \dfrac{\pi}{2}$.
The points $(\pi - x, \sin(\pi - x)) = (\pi - x, \sin x)$ and $(\pi + x, \sin(\pi + x)) = (\pi + x, -\sin x)$ have opposite $y$-coordinates, and $\dfrac{(\pi - x) + (\pi + x)}{2} = \pi$.
The points $(x, \cos x)$ and $(\pi - x, \cos(\pi - x)) = (\pi - x, -\cos x)$ have opposite $y$-coordinates, and $\dfrac{x + (\pi - x)}{2} = \dfrac{\pi}{2}$.
The point $(x, \sin x) = \left(x, \cos\left(x - \frac{\pi}{2}\right)\right)$ of the sine curve is the point $\left(x - \frac{\pi}{2}, \cos\left(x - \frac{\pi}{2}\right)\right)$ of the cosine curve moved $\dfrac{\pi}{2}$ to the right; and $\cos\left(x - \frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2} - x\right) = \sin x$, since cosine is even.

Example

Without a calculator, compare:

$\cos 1$ and $\cos 2$;
$\sin 0.5$ and $\sin 1$;
$\sin 1$ and $\sin 2$.

Answer

$1$ and $2$ lie in $\left[0, \pi\right]$, where cosine is decreasing, and $1 \lt 2$: so $\cos 1 \gt \cos 2$.
$0.5$ and $1$ lie in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (as $\frac{\pi}{2} \approx 1.57$), where sine is increasing, and $0.5 \lt 1$: so $\sin 0.5 \lt \sin 1$.
$2 \gt \frac{\pi}{2}$, so $1$ and $2$ are not in the same interval of variation. The symmetry about $x = \frac{\pi}{2}$ brings $2$ back: $\sin 2 = \sin(\pi - 2)$, with $\pi - 2 \gt 1$ since $\pi \gt 3$, and $\pi - 2 \lt \frac{\pi}{2}$ since $\pi \lt 4$. Now $1$ and $\pi - 2$ lie in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $1 \lt \pi - 2$: so $\sin 1 \lt \sin(\pi - 2) = \sin 2$.

Skills to practice

Using the Variations and Symmetries of the Curves

III General Sine and Cosine Functions

III.1 The Functions $y = a\sin\big(b(x - c)\big) + d$

Recappreviously studied — click to reread

Now that we are familiar with the graphs of $y = \sin x$ and $y = \cos x$, we can use transformations to graph more complicated trigonometric functions.

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Definition — General Sine and Cosine Function

For functions of the form $y = a\sin(b(x-c)) + d$ and $y = a\cos(b(x-c)) + d$, with $a \neq 0$ and $b \neq 0$:

$|a|$ is the amplitude.
$\frac{2\pi}{|b|}$ is the period ($x$ in radians; with $x$ in degrees, the period is $\frac{360^\circ}{|b|}$).
$c$ is the phase shift (horizontal translation).
$d$ is the vertical shift (the principal axis is $y=d$).

Proposition — Transformations of Trigonometric Functions

For real parameters $a,b,c,d$ with $b\neq 0$, the graph of $y = a\sin\!\big(b(x-c)\big) + d$ is obtained from the graph of $y=\sin x$ by:

a vertical stretch by factor $|a|$; if $a<0$, there is also a reflection in the $x$-axis,
a horizontal stretch by factor $\dfrac{1}{|b|}$; if $b<0$, there is also a reflection in the $y$-axis,
a horizontal translation by $c$ units (to the right if $c>0$, left if $c<0$) and a vertical translation by $d$ units (up if $d>0$, down if $d<0$).

Example

Sketch the graph of $y=\sqrt{2} \sin x-1$ for $0 \leq x \leq 2 \pi$.

Answer

The graph of $y=\sqrt{2} \sin x$ is the vertical stretch of the graph of $y=\sin x$ with scale factor $\sqrt{2}$.
The graph of $y=\sqrt{2} \sin x-1$ is the vertical translation of the graph of $y=\sqrt{2}\sin x$ by $-1$.

Skills to practice

Finding the Equation from a Graph
Identifying Properties from an Equation

IV Modelling Periodic Behaviour

IV.1 Sinusoidal Models

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
$T(d)$ is the temperature of a city $d$ 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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Definition — Sine and Cosine Models

An Sine/Cosine Model is a function of the form $$y = a\sin(b(x-c)) + d\text{ or }y = a\cos(b(x-c)) + d$$

$|a|$ is the amplitude.
$\frac{2\pi}{|b|}$ (or $\frac{360}{|b|}$) is the period.
$c$ is the phase shift (horizontal translation).
$d$ is the vertical shift (the midline is $y=d$).

Example

The water depth, $D$, in metres, at a port can be modelled by the function$$ D(t) = 3 \sin\left( \frac{\pi}{12} (t - 2) \right) + 12, \quad 0 \leq t < 24 $$ where $t$ is the elapsed time, in hours, since midnight.

Write down the depth of the water at midnight.
Find the minimum water depth and the time it occurs.
Calculate how long the water depth is at least 13.5 metres each day.

Answer

At midnight ($t=0$): $$ D(0) = 3 \sin\left( \frac{\pi}{12} (0 - 2) \right) + 12 = 3 \sin\!\left(-\frac{\pi}{6}\right) + 12. $$ $$ D(0) = 3(-0.5) + 12 = 10.5 \text{ metres}. $$
Minimum depth: The sine function ranges from $-1$ to $1$, so the minimum occurs when $\sin(\dots)=-1$: $$ D_{\min} = 3(-1) + 12 = 9 \text{ metres}. $$ This occurs when the argument equals $-\frac{\pi}{2} + 2\pi k$. In the interval $0\le t<24$, we may use $\frac{3\pi}{2}$: $$ \frac{\pi}{12}(t-2) = \frac{3\pi}{2} \implies t-2 = 18 \implies t = 20 \text{ hours (8 PM)}. $$
Depth $\ge 13.5$: $$ 3 \sin\left( \frac{\pi}{12} (t - 2) \right) + 12 \ge 13.5 \iff \sin\left( \frac{\pi}{12} (t - 2) \right) \ge 0.5. $$ The equality $\sin\theta=0.5$ occurs at $\theta=\frac{\pi}{6}$ and $\theta=\frac{5\pi}{6}$ (within one period).
- $\frac{\pi}{12}(t-2) = \frac{\pi}{6} \implies t-2 = 2 \implies t=4$.
- $\frac{\pi}{12}(t-2) = \frac{5\pi}{6} \implies t-2 = 10 \implies t=12$.
Therefore, $D(t)\ge 13.5$ for $4 \le t \le 12$, which lasts $12-4=8$ hours.

Skills to practice

Using Sinusoidal Models with Technology

IV.2 Fitting a Sine Model to Data

Method — Modelling Periodic Data with a Sine Function

To find a sinusoidal model of the form $y = a\sin(b(x-c)) + d$ that fits a set of periodic data:

Find the vertical shift, $d$: Calculate the principal axis, which is the average of the maximum and minimum values. $$d = \frac{\text{max value} + \text{min value}}{2}$$
Find the amplitude, $a$: Calculate half the distance between the maximum and minimum values. $$a = \frac{\text{max value} - \text{min value}}{2}$$
Find the period and the parameter $b$: Determine the period, $P$, which is the length of one full cycle of the data. Use the formula $P = \frac{2\pi}{b}$ to solve for $b$. $$b = \frac{2\pi}{P}$$
Find the phase shift, $c$: Substitute a data point $(x,y)$ into the equation $y = a\sin(b(x-c))+d$ and solve for the horizontal shift $c$. It is often easiest to choose a point where the function crosses its principal axis, as this corresponds to a point where the sine function is zero.

Example

The mean monthly maximum temperatures for Cape Town, South Africa are shown below:

Month ($t$)	1	2	3	4	5	6	7	8	9	10	11	12
Temp ($T$)	28	27	25.5	22	18.5	16	15	16	18	21.5	24	26

Find a sinusoidal function of the form $T = a\sin(b(t-c))+d$ to model the temperature.

Answer

Find Vertical Shift ($d$): The max value is 28 and the min value is 15. $$ d = \frac{28+15}{2} = \frac{43}{2} = 21.5 $$
Find Amplitude ($a$): $$ a = \frac{28-15}{2} = \frac{13}{2} = 6.5 $$
Find Period ($b$): The data repeats every 12 months, so the period $P=12$. $$ b = \frac{2\pi}{P} = \frac{2\pi}{12} = \frac{\pi}{6} $$
Find Phase Shift ($c$): Our model so far is $T = 6.5\sin(\frac{\pi}{6}(t-c)) + 21.5$. A standard sine function starts at its principal axis and goes up. From the data, the temperature is 21.5 (the principal axis value) at $t=10$ and is increasing. We can therefore choose the phase shift to be $c=10$.
Final Model: $$ T(t) = 6.5\sin\left(\frac{\pi}{6}(t-10)\right) + 21.5 $$

Skills to practice

Building and Using Sinusoidal Models

V The Tangent Function

V.1 The General Tangent Function

Proposition — Graph of the Tangent Function

The graph of $y = \tan(x) = \frac{\sin x}{\cos x}$ has:

Domain: $\{x \in \mathbb{R} \mid x \neq \frac{\pi}{2} + k\pi,\forall k\in\mathbb{Z}\}$ (all real numbers $x$ except those for which $\cos x = 0$)
Period: $\pi$.
Vertical Asymptotes: at $x = \frac{\pi}{2} + k\pi$ for any integer $k$.
Range: $\mathbb{R}$.

Proposition — General Tangent Function

Let $a$, $b$, $c$ and $d$ be real numbers, with $a \neq 0$ and $b \neq 0$, and let $f(x) = a\tan\big(b(x - c)\big) + d$. The function $f$:

has period $\dfrac{\pi}{|b|}$;
has the vertical asymptotes $x = c + \dfrac{1}{b}\left(\dfrac{\pi}{2} + k\pi\right)$, for every integer $k$: the values of $x$ for which $\cos\big(b(x - c)\big) = 0$;
meets the line $y = d$ at $x = c + \dfrac{k\pi}{b}$, for every integer $k$, halfway between two consecutive asymptotes: this line through the centres of the branches is called its principal axis;
has no amplitude: its range is $\mathbb{R}$.

Its graph is the graph of $y = \tan x$ stretched vertically by the factor $|a|$ and horizontally by the factor $\dfrac{1}{|b|}$, reflected in the horizontal axis if $a < 0$ and in the vertical axis if $b < 0$, then translated by $c$ horizontally and by $d$ vertically.

Proof

Write $u = b(x - c)$, so that $f(x) = a\tan u + d$. The tangent function has period $\pi$, vertical asymptotes $u = \dfrac{\pi}{2} + k\pi$ and range $\mathbb{R}$ (Proposition “Graph of the Tangent Function”); moreover $\tan u = 0$ exactly when $\sin u = 0$, that is when $u$ is a multiple of $\pi$ (the points $(1, 0)$ and $(-1, 0)$ of the unit circle).

Period. Since $b \times \dfrac{\pi}{|b|}$ equals $\pi$ or $-\pi$, for every $x$ where $f$ is defined, $$\begin{aligned}[t] f\left(x + \frac{\pi}{|b|}\right) &= a\tan\big(b(x - c) \pm \pi\big) + d && \text{(expand)}\\ &= a\tan\big(b(x - c)\big) + d && \text{(period } \pi \text{ of } \tan\text{)}\\ &= f(x). \end{aligned}$$ No smaller positive number works: if $p > 0$ and $f(x + p) = f(x)$ for every $x$, then at $x = c$, $a\tan(bp) + d = a\tan 0 + d = d$, so $\tan(bp) = 0$ since $a \neq 0$. Hence $bp$ is a multiple of $\pi$, nonzero since $bp \neq 0$, and $|b|p \geq \pi$, that is $p \geq \dfrac{\pi}{|b|}$.
Asymptotes. $f(x)$ is defined exactly when $\cos u \neq 0$. Solving $b(x - c) = \dfrac{\pi}{2} + k\pi$ for $x$ gives $x = c + \dfrac{1}{b}\left(\dfrac{\pi}{2} + k\pi\right)$. As $x$ approaches one of these values, $\tan u$ takes arbitrarily large values in absolute value, and so does $f(x)$, since $a \neq 0$.
Principal axis. $$\begin{aligned}[t] f(x) = d &\iff a\tan u = 0 && \text{(subtract } d\text{)}\\ &\iff \tan u = 0 && (a \neq 0)\\ &\iff u = k\pi && (k \text{ an integer})\\ &\iff x = c + \frac{k\pi}{b}. && \text{(divide by } b\text{, add } c\text{)} \end{aligned}$$ Two consecutive asymptotes correspond to $u = \dfrac{\pi}{2} + k\pi$ and $u = \dfrac{\pi}{2} + (k + 1)\pi$; halfway between them, $u = (k + 1)\pi$, one of the values found.
Range. For every real number $y$, the number $\dfrac{y - d}{a}$ is a value of the tangent function, $\tan u_0$ for some $u_0$; then $y = a\tan u_0 + d$ is a value of $f$.

The description of the graph applies to $y = \tan x$ the transformations of graphs (stretches, reflections, translations), as for the sine and the cosine.

Example

For $y = \tan(3x)$, state the period, then find the vertical asymptotes and the points where the graph meets its principal axis for $0 \le x \le \pi$. Sketch the graph for $0 \le x \le \pi$.
For $y = 2\tan\left(x - \dfrac{\pi}{4}\right) + 1$, state the period and the principal axis, and write the equations of the vertical asymptotes.

Answer

Here $a = 1$, $b = 3$, $c = 0$ and $d = 0$. The period is $\dfrac{\pi}{|b|} = \dfrac{\pi}{3}$. The asymptotes are the values of $x$ for which $\cos(3x) = 0$: $$\begin{aligned}[t] 3x = \frac{\pi}{2} + k\pi &\iff x = \frac{\pi}{6} + \frac{k\pi}{3}. && \text{(divide by } 3\text{)} \end{aligned}$$ For $0 \le x \le \pi$, the values $k = 0$, $1$, $2$ give the asymptotes $x = \dfrac{\pi}{6}$, $x = \dfrac{\pi}{2}$ and $x = \dfrac{5\pi}{6}$. The principal axis is $y = 0$, met at $x = \dfrac{k\pi}{3}$: at $x = 0$, $\dfrac{\pi}{3}$, $\dfrac{2\pi}{3}$ and $\pi$, halfway between the asymptotes. Since $a > 0$, each branch increases.
Here $a = 2$, $b = 1$, $c = \dfrac{\pi}{4}$ and $d = 1$. The period is $\dfrac{\pi}{1} = \pi$ and the principal axis is $y = 1$. The asymptotes: $$\begin{aligned}[t] x - \frac{\pi}{4} = \frac{\pi}{2} + k\pi &\iff x = \frac{3\pi}{4} + k\pi, && \text{(add } \tfrac{\pi}{4}\text{)} \end{aligned}$$ for every integer $k$. The graph meets its principal axis $y = 1$ at $x = \dfrac{\pi}{4} + k\pi$.

Skills to practice

Finding the Period and Asymptotes of a Tangent Function
Sketching Tangent Functions

VI Trigonometric Equations

VI.1 Solving Trigonometric Equations

Recappreviously studied — click to reread

Method — Solving a Trigonometric Equation

Use identities to simplify the equation into a form involving a single trigonometric function, e.g., $\sin(X)=k$.
Find one solution $X_0$ with an inverse function, for example $X_0 = \arcsin(k)$ (the key $\sin^{-1}$ of the calculator) for $\sin X = k$ with $-1 \leq k \leq 1$. This principal value lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ (between $-90^\circ$ and $90^\circ$), so it is negative when $k < 0$: for example $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$, which is not in $[0, 2\pi)$.
Use the symmetry and periodicity of the function to find all solutions within one period.
Add whole multiples of the period — $2k\pi$ (that is $k \times 360^\circ$) for the sine and the cosine, $k\pi$ (that is $k \times 180^\circ$) for the tangent, $k$ an integer — to find the general solution or all the solutions in a given interval. A negative principal value gives a solution in $[0, 2\pi)$ once $2\pi$ is added: $-\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}$.

Skills to practice

Solving Basic Trigonometric Equations without Technology
Solving Equations with Transformed Arguments

VI.2 Solving with the Unit Circle

How Many Solutions Does a Trigonometric Equation Have?

On the unit circle, one reads at once that $\dfrac{\pi}{3}$ is a solution of $\cos(x) = \dfrac{1}{2}$. Is it the only one? An equation of this kind usually has infinitely many solutions, and a question almost never asks for one of them: it asks for all those belonging to a given interval. Two things are therefore needed, and the property below provides both: a way of describing the whole set of solutions, and a way of extracting from it those of the interval asked for.

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Proposition — Solving $\cos(x) \equal \cos(a)$ and $\sin(x) \equal \sin(a)$

Let $a$ be a real number. For every real number $x$: $$\cos(x) = \cos(a) \iff x = a + 2k\pi \ \text{ or } \ x = -a + 2k\pi, \qquad k \in \mathbb{Z};$$ $$\sin(x) = \sin(a) \iff x = a + 2k\pi \ \text{ or } \ x = \pi - a + 2k\pi, \qquad k \in \mathbb{Z}.$$ The unit circle explains both. The points with the same abscissa $\cos(a)$ are symmetric about the horizontal axis, which gives $a$ and $-a$:

The points with the same ordinate $\sin(a)$ are symmetric about the vertical axis, which gives $a$ and $\pi - a$:

In each case, any whole number of turns may be added.

Method — Solving a Trigonometric Equation Using the Unit Circle

This method applies when the equation can be written with a remarkable value on the right-hand side. For a value that is not remarkable, the calculator gives an approximation of one solution and the circle then gives the others.

Write the equation in the form $\cos(X) = \cos(a)$ or $\sin(X) = \sin(a)$, where $a$ is read from the unit circle.
Write the two families of solutions in $X$: $X = a + 2k\pi$ or $X = -a + 2k\pi$ for the cosine, $X = a + 2k\pi$ or $X = \pi - a + 2k\pi$ for the sine ($k \in \mathbb{Z}$).
Go back to $x$ if $X$ was an expression such as $2x$ or $3x - \frac{\pi}{4}$.
Select the integers $k$ for which the solution belongs to the interval required by the question, then list the solutions.

Example

Solve the equation $2\cos(2x) = 1$ on the interval $[0, 2\pi)$.

Answer

For every real number $x$, denoting by $k$ an integer: $$\begin{aligned}[t] 2\cos(2x) = 1 &\iff \cos(2x) = \dfrac{1}{2} &&\text{(divide by 2)} \\ &\iff \cos(2x) = \cos\left(\dfrac{\pi}{3}\right) &&\text{(remarkable value)} \\ &\iff 2x = \dfrac{\pi}{3} + 2k\pi \ \text{ or } \ 2x = -\dfrac{\pi}{3} + 2k\pi &&\text{(previous proposition)} \\ &\iff x = \dfrac{\pi}{6} + k\pi \ \text{ or } \ x = -\dfrac{\pi}{6} + k\pi \end{aligned}$$ It remains to select the integers $k$ giving a solution in $[0, 2\pi)$. For the first family, $k = -1$ gives $-\dfrac{5\pi}{6} < 0$, $k = 0$ gives $\dfrac{\pi}{6}$, $k = 1$ gives $\dfrac{7\pi}{6}$, and $k = 2$ gives $\dfrac{13\pi}{6} > 2\pi$. For the second family, $k = 0$ gives $-\dfrac{\pi}{6} < 0$, $k = 1$ gives $\dfrac{5\pi}{6}$, $k = 2$ gives $\dfrac{11\pi}{6}$, and $k = 3$ gives $\dfrac{17\pi}{6} > 2\pi$. The solution set is $$\left\{\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}\right\}.$$

Skills to practice

Solving an Equation with the Two Families of Solutions

VI.3 Equations of Quadratic Type

Method — Solving a Trigonometric Equation of Quadratic Type

An equation such as $2\cos^2 x - \cos x - 1 = 0$ is a quadratic equation in the unknown $u = \cos x$; the same holds with $\sin x$ or $\tan x$.

Bring every term to one side, and do not divide by $\sin x$, $\cos x$ or $\tan x$ when this number can be $0$: the solutions for which it is $0$ would be lost. Factorise instead.
Solve the quadratic equation in $u$: by factorising and the Null Factor Law, or with the quadratic formula.
Keep the admissible values: $\sin x$ and $\cos x$ lie between $-1$ and $1$, so a value of $u$ outside $[-1, 1]$ gives no solution; $\tan x$ takes every real value.
Solve each basic equation ($\cos x = u_1$, $\cos x = u_2$, …) on the interval, with the unit circle when the value is remarkable and with technology otherwise, then list all the solutions in increasing order.

Example

Solve each equation for $0 \le x \le 2\pi$, giving exact values.

$2\cos^2 x - \cos x - 1 = 0$;
$2\sin^2 x = \sqrt{3}\sin x$;
$2\sin^2 x + 5\sin x - 3 = 0$.

Answer

With $u = \cos x$, the left side is $2u^2 - u - 1 = (2u + 1)(u - 1)$ (check by expanding: $2u^2 - 2u + u - 1$). $$\begin{aligned}[t] 2\cos^2 x - \cos x - 1 = 0 &\iff (2\cos x + 1)(\cos x - 1) = 0 && \text{(factorise)}\\ &\iff \cos x = -\frac{1}{2} \ \text{ or } \ \cos x = 1. && \text{(Null Factor Law)} \end{aligned}$$ Both values lie between $-1$ and $1$. On the unit circle, $\cos x = -\dfrac{1}{2}$ for $x = \dfrac{2\pi}{3}$ and $x = \dfrac{4\pi}{3}$, and $\cos x = 1$ for $x = 0$ and $x = 2\pi$. The solutions are $0$, $\dfrac{2\pi}{3}$, $\dfrac{4\pi}{3}$ and $2\pi$.
The equation is not divided by $\sin x$, which can be $0$: $$\begin{aligned}[t] 2\sin^2 x = \sqrt{3}\sin x &\iff 2\sin^2 x - \sqrt{3}\sin x = 0 && \text{(subtract } \sqrt{3}\sin x\text{)}\\ &\iff \sin x\,(2\sin x - \sqrt{3}) = 0 && \text{(common factor)}\\ &\iff \sin x = 0 \ \text{ or } \ \sin x = \frac{\sqrt{3}}{2}. && \text{(Null Factor Law)} \end{aligned}$$ $\sin x = 0$ for $x = 0$, $\pi$ and $2\pi$; $\sin x = \dfrac{\sqrt{3}}{2}$ for $x = \dfrac{\pi}{3}$ and $x = \dfrac{2\pi}{3}$. The solutions are $0$, $\dfrac{\pi}{3}$, $\dfrac{2\pi}{3}$, $\pi$ and $2\pi$. Dividing by $\sin x$ would have kept only $\dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$.
With $u = \sin x$, the left side is $2u^2 + 5u - 3 = (2u - 1)(u + 3)$ (check: $2u^2 + 6u - u - 3$). $$\begin{aligned}[t] 2\sin^2 x + 5\sin x - 3 = 0 &\iff (2\sin x - 1)(\sin x + 3) = 0 && \text{(factorise)}\\ &\iff \sin x = \frac{1}{2} \ \text{ or } \ \sin x = -3. && \text{(Null Factor Law)} \end{aligned}$$ Since $-3 < -1$, the equation $\sin x = -3$ has no solution. $\sin x = \dfrac{1}{2}$ for $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$: these are the solutions.

Skills to practice

Solving Equations of Quadratic Type without Technology

VI.4 Solving Trigonometric Equations with Technology

The unit circle solves $\cos x = \dfrac{1}{2}$, because $\dfrac{1}{2}$ is the cosine of a remarkable angle. In $3\cos x = 1$, that is $\cos x = \dfrac{1}{3}$, the value $\dfrac{1}{3}$ is the cosine of no remarkable angle, and none of the exact methods solves $2\sin x = x - 1$, where the unknown appears both inside and outside the sine. A graphing calculator approximates such solutions. For $3\cos x = 1$, the period and the symmetries of the cosine tell how many solutions to look for and check the values read on the screen; for $2\sin x = x - 1$, the screen shows the solutions, and a comparison of the two sides, interval by interval, confirms that none is missed.

Report an error

Method — Solving a Trigonometric Equation with Technology

The method “Solving an Equation Graphically with Technology” applies, with the following steps specific to trigonometric functions.

Set the angle mode (Method “Setting the Calculator Mode”): radians when the interval is written with real numbers or with $\pi$, degrees when it is written in degrees.
Enter each side as a function, $Y_1$ = left side and $Y_2$ = right side, and take the window from the interval: $x$ runs over exactly the interval given, and $y$ covers the values of both sides (the values of $a\sin(bx) + d$ lie between $d - |a|$ and $d + |a|$; a tangent has no bounds, and a range around the value sought is enough).
Count the points of intersection before reading them. For $\sin(bx) = k$ or $\cos(bx) = k$ with $-1 < k < 1$, expect two solutions in each period $\dfrac{2\pi}{|b|}$ taken with one end only; for $\tan(bx) = k$, one in each period $\dfrac{\pi}{|b|}$. Then check whether the two ends of the interval are themselves solutions: $\sin x = 0$ has three solutions for $0 \le x \le 2\pi$, namely $0$, $\pi$ and $2\pi$.
With the intersection tool, read the $x$-coordinate of each point, and round as the question asks (to 3 significant figures if it says nothing). For a tangent, the screen may join two branches by a vertical segment at an asymptote: a crossing there is not a solution, since the tangent is not defined at that value.
Check with the symmetries: the solutions of $\cos x = k$ go in pairs $x_0$ and $2\pi - x_0$ (the cosine is even, of period $2\pi$), those of $\sin x = k$ in pairs $x_0$ and $\pi - x_0$ (as $\sin(\pi - x) = \sin x$), the two values coinciding, up to a period, when $k = 1$ or $k = -1$; adding or subtracting a period gives the others, and only the values that lie in the interval are kept.

Example

Solve each equation with technology, giving the solutions to 3 significant figures.

$3\cos x = 1$, for $0 \le x \le 4\pi$;
$2\sin x = x - 1$, for $0 \le x \le 2\pi$.

Answer

The calculator is set to radians. The equation is $\cos x = \dfrac{1}{3}$, with $-1 < \dfrac{1}{3} < 1$, and the interval $0 \le x \le 4\pi$ covers two periods of the cosine: four solutions are expected. With $Y_1 = 3\cos x$ and $Y_2 = 1$, in the window $0 \le x \le 4\pi$ (about $12.6$), $-4 \le y \le 4$, the curves meet four times, and the intersection tool displays $X = 1.2309594$, $5.0522259$, $7.5141447$ and $11.3354112$. Check with the symmetries: $2\pi - 1.2309594 \approx 5.0522259$, $2\pi + 1.2309594 \approx 7.5141447$ and $4\pi - 1.2309594 \approx 11.3354112$. The solutions are $$x \approx 1.23, \quad x \approx 5.05, \quad x \approx 7.51, \quad x \approx 11.3.$$
The unknown appears inside and outside the sine: none of the exact methods applies, and technology approximates the solution. With $Y_1 = 2\sin x$ and $Y_2 = x - 1$, in the window $0 \le x \le 2\pi$, $-3 \le y \le 6$, the curves meet once, and the intersection tool displays $X = 2.3800613$ and $Y = 1.3800613$. There is no other solution, as the two sides show:
- for $0 \le x \le 1$: $x - 1 \le 0 \le 2\sin x$, and the two sides are never both $0$;
- for $1 < x \le \dfrac{\pi}{2}$: the sine increases, so $2\sin x > 2\sin 1 > 1.6 > \dfrac{\pi}{2} - 1 \ge x - 1$;
- for $\dfrac{\pi}{2} \le x \le \pi$: $2\sin x$ decreases while $x - 1$ increases, so the curves meet at most once, at the point displayed;
- for $\pi \le x \le 2\pi$: $2\sin x \le 0 < x - 1$.
The solution is $x \approx 2.38$. Check: $2\sin(2.38) \approx 1.380$ and $2.38 - 1 = 1.38$.

Skills to practice

Counting the Solutions on an Interval
Solving Equations with the Intersection Tool
Solving Equations in Context with Technology

VII Going further

Skills to practice

Reasoning and Problem-Solving
Solving Problems
Quiz

\(x\)	0	\(\dfrac{\pi}{6}\)	\(\dfrac{\pi}{4}\)	\(\dfrac{\pi}{3}\)	\(\dfrac{\pi}{2}\)	\(\dfrac{2 \pi}{3}\)	\(\dfrac{3 \pi}{4}\)	\(\dfrac{5 \pi}{6}\)	\(\pi\)
\(\sin (x)\)

\(x\)	0	\(\dfrac{\pi}{6}\)	\(\dfrac{\pi}{4}\)	\(\dfrac{\pi}{3}\)	\(\dfrac{\pi}{2}\)	\(\dfrac{2 \pi}{3}\)	\(\dfrac{3 \pi}{4}\)	\(\dfrac{5 \pi}{6}\)	\(\pi\)
\(\sin (x)\)	0	\(\dfrac{1}{2}\)	\(\dfrac{\sqrt{2}}{2}\)	\(\dfrac{\sqrt{3}}{2}\)	1	\(\dfrac{\sqrt{3}}{2}\)	\(\dfrac{\sqrt{2}}{2}\)	\(\dfrac{1}{2}\)	\(0\)

\(x\)	0	\(\dfrac{\pi}{6}\)	\(\dfrac{\pi}{4}\)	\(\dfrac{\pi}{3}\)	\(\dfrac{\pi}{2}\)	\(\dfrac{2 \pi}{3}\)	\(\dfrac{3 \pi}{4}\)	\(\dfrac{5 \pi}{6}\)	\(\pi\)
\(\cos (x)\)

\(x\)	\(0\)	\(\dfrac{\pi}{6}\)	\(\dfrac{\pi}{4}\)	\(\dfrac{\pi}{3}\)	\(\dfrac{\pi}{2}\)	\(\dfrac{2 \pi}{3}\)	\(\dfrac{3 \pi}{4}\)	\(\dfrac{5 \pi}{6}\)	\(\pi\)
\(\cos (x)\)	\(1\)	\(\dfrac{\sqrt{3}}{2}\)	\(\dfrac{\sqrt{2}}{2}\)	\(\dfrac{1}{2}\)	\(0\)	\(-\dfrac{1}{2}\)	\(-\dfrac{\sqrt{2}}{2} \)	\(-\dfrac{\sqrt{3}}{2}\)	\(-1\)

Month (\(t\))	1	2	3	4	5	6	7	8	9	10	11	12
Temp (\(T\))	28	27	25.5	22	18.5	16	15	16	18	21.5	24	26