International Baccalaureate · DP 2: Analysis and Approaches SL

Trigonometric Functions

⌚ ~122 min ▢ 13 blocks ✓ 58 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.

Sine Regression on a Calculator

A graphing calculator can also fit a sine model to a set of data by sine regression, in radian mode. It returns the model in the form $y = a\sin(bx + k) + d$. The constant written $k$ here is the one most calculators call $c$, and it is not the horizontal translation $c$ of $y = a\sin\big(b(x - c)\big) + d$. To go from one form to the other, factorise by $b$ inside the sine: $$a\sin(bx + k) + d = a\sin\left(b\left(x + \frac{k}{b}\right)\right) + d, \qquad \text{so} \qquad c = -\frac{k}{b}.$$ For instance, a regression that returns $y = 6.5\sin(0.524x - 5.24) + 21.5$ gives $c = -\dfrac{-5.24}{0.524} = 10$: the model is $y = 6.5\sin\big(0.524(x - 10)\big) + 21.5$, with $b \approx 0.524 \approx \dfrac{\pi}{6}$, a period of about $12$. The values displayed are rounded, and a calculator may return another form of the same curve — a value of $k$ that differs by a multiple of $2\pi$, or the opposite of $a$ with $\pi$ added to $k$, since $a\sin\theta = -a\sin(\theta + \pi)$. The model is read from the values, whatever the form.

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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 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