Lesson 3 of 5Trigonometric Functions

Unit 4 · Lesson 3 of 5

Trigonometric Graphs and Transformations

Unroll the unit circle onto an axis and you get the sine, cosine, and tangent graphs — then the familiar four transformation parameters reshape them, with period now measured in multiples of pi.

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Key things to remember

1 $y = sin x$ and $y = cos x$ : period $2 π$ , amplitude $1$ , axis $y = 0$ , range $[- 1, 1]$ , domain all real numbers.
2 $y = tan x$ : period $π$ , range all real numbers, vertical asymptotes at $x = \frac{π}{2} + nπ$ , zeros at $x = nπ$ .
3In $y = a sin (k (x - d)) + c$ : amplitude $= ∣ a ∣$ , period $= \frac{2 π}{∣ k ∣}$ , phase shift $= d$ (right if positive), axis $= y = c$ .
4Maximum $= c + ∣ a ∣$ and minimum $= c - ∣ a ∣$ .
5From a graph: $c = \frac{max + min}{2}$ , $a = \frac{max - min}{2}$ , $k = \frac{2 π}{period}$ .
6Always factor $k$ out of the bracket before reading the phase shift.

Take the unit circle and imagine unrolling it: walk counterclockwise around the circle and plot the height ( $y$ -coordinate) against the distance travelled. The height rises to $1$ , falls back through $0$ to $- 1$ , and returns — and because the walk repeats after $2 π$ of arc, so does the graph. That unrolled picture is $y = sin x$ , and the same walk plotting the $x$ -coordinate gives $y = cos x$ .

The parent graphs in radians

In MCR3U these curves repeated every $360°$ ; nothing about their shape has changed, only the label on the horizontal axis. The period is now $2 π$ , the interesting landmarks sit at multiples of $\frac{π}{2}$ , and the domain of both is all real numbers while the range of both is $- 1 \leq y \leq 1$ .

Fig 1One full period of each parent curve, marked at the quarter points.

Feature	$y = sin x$	$y = cos x$
Value at $x = 0$	$0$	$1$
Maximum of $1$ at	$x = \frac{π}{2}$	$x = 0$
Zeros at	$x = nπ$	$x = \frac{π}{2} + nπ$
Minimum of $- 1$ at	$x = \frac{3 π}{2}$	$x = π$
Period	$2 π$	$2 π$
Amplitude / axis	$1$ / $y = 0$	$1$ / $y = 0$

Here n is any integer. The two curves are the same shape shifted horizontally by a quarter period.

VideoAn Ontario teacher graphs the parent trig functions with radian labels on the horizontal axis.Watch on YouTube

Why the tangent graph has asymptotes

Tangent is a quotient, $tan x = \frac{sin x}{cos x}$ , so it is undefined wherever the denominator is zero — that is, at $x = \frac{π}{2} + nπ$ . Approaching those angles, $cos x$ shrinks toward $0$ while $sin x$ stays near $\pm 1$ , so the quotient blows up: a vertical asymptote. Between consecutive asymptotes the curve sweeps from $- \infty$ all the way up to $+ \infty$ , crossing zero wherever $sin x = 0$ (at $x = nπ$ ). Crucially, tangent repeats every $π$ , not every $2 π$ — its period is HALF that of sine and cosine, and its range is all real numbers, so tangent has no amplitude at all.

Transformations

y = a sin (k (x - d)) + c y = a cos (k (x - d)) + c

$a$ — vertical stretch. The amplitude becomes $∣ a ∣$ , so the maximum is $c + ∣ a ∣$ and the minimum is $c - ∣ a ∣$ . A negative $a$ reflects the curve in its axis (flips it upside down) without changing the amplitude.
$k$ — horizontal stretch or compression. The period becomes $\frac{2 π}{∣ k ∣}$ , so a bigger $∣ k ∣$ squeezes more cycles into the same space. For tangent the period is $\frac{π}{∣ k ∣}$ instead, because tangent starts with a period of $π$ .
$d$ — horizontal (phase) shift. The graph slides RIGHT by $d$ when $d > 0$ and left when $d < 0$ .
$c$ — vertical shift. The axis (midline) moves to the line $y = c$ .

Fig 2The same parent sine curve under a vertical stretch, a horizontal compression, a phase shift, and a vertical shift.

Factor out k before reading d

The phase shift is only visible once the coefficient of x is factored out of the bracket. In y = sin(2x - pi/3) the shift is NOT pi/3. Rewrite it as y = sin(2(x - pi/6)) first, and the shift is pi/6 to the right. Forgetting to factor is the single most common error on this topic.

VideoAn Ontario Grade 12 lesson on applying the transformation parameters a, k, d and c to trig functions in radians.Watch on YouTube

Reading an equation off a graph, and modelling real data

Fig 3Maximum, minimum, axis, amplitude, and period — the five measurements you take off any sinusoid.

Read the maximum and minimum. Then $c = \frac{max + min}{2}$ is the axis and $a = \frac{max - min}{2}$ is the amplitude.
Measure the period as the horizontal distance between two consecutive maxima (or two consecutive minima), then solve $k = \frac{2 π}{period}$ .
Choose the parent function from the starting behaviour: at a maximum use $+ cos$ , at a minimum use $- cos$ , and at the axis heading upward use $+ sin$ . Picking well usually makes $d = 0$ .
If the graph does not start conveniently, keep cosine and set $d$ to the $x$ -value of the first maximum.
For real data (tides, daylight hours, temperature, a Ferris wheel), run the same four steps on the numbers in the table, then always check your equation against a known data point before trusting it.

Loading interactive…

Move the four sliders and watch amplitude, period, phase shift, and axis respond.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

State the amplitude, period, phase shift, axis, maximum, and minimum of

y = 3 sin (2 (x - \frac{π}{4})) + 1

Example 2

Describe how

y = - 2 cos (\frac{x}{3}) + 5

differs from

y = cos x

Example 3

A sinusoidal graph has a maximum of

7

, a minimum of

- 1

, and its first maximum at

x = \frac{π}{2}

. Consecutive maxima are

4 π

apart. Write an equation for it.

Example 4

The water depth at a dock varies from

1.5

m at low tide to

5.5

m at high tide on a

12

-hour cycle, with high tide at

t = 0

. Write a model

D (t)

for the depth in metres after

t

hours, then find the depth at

t = 3

Where marks get lost

Using $\frac{360°}{∣ k ∣}$ for the period out of habit. In radians the period of sine and cosine is $\frac{2 π}{∣ k ∣}$ .
Reading the phase shift straight out of an unfactored bracket — $y = sin (2 x - π /3)$ shifts by $\frac{π}{6}$ , not $\frac{π}{3}$ , because $k$ must be factored out first.
Giving tangent a period of $2 π$ or an amplitude. Tangent repeats every $π$ , and since its range is all real numbers it has no maximum, no minimum, and no amplitude.
Reading the sign of $d$ backwards: $(x - d)$ with a positive $d$ shifts RIGHT, so $y = cos (x + π /4)$ means $d = - \frac{π}{4}$ , a shift LEFT.
Confusing amplitude with the full vertical spread. Amplitude is HALF the distance from minimum to maximum, so a curve running from $- 1$ to $7$ has amplitude $4$ , not $8$ .

Flashcards 8

Period of

y = a sin (k (x - d)) + c

Period of

y = tan (k x)

Asymptotes of

y = tan x

Maximum and minimum of a sinusoid

+ 4 more

Practice

Find the period of

y = sin (3 x)

. Give the decimal value, rounded to 2 decimals.

Find the amplitude of

y = - 6 cos (2 (x - \frac{π}{3})) + 1

What is the equation of the axis of

y = - 6 cos (2 (x - \frac{π}{3})) + 1

Find the period of

y = tan (2 x)

. Give the decimal value, rounded to 3 decimals.

Does the graph of

y = sin (x - \frac{π}{3})

shift left or right compared with

y = sin x

For

y = 5 sin (2 (x - \frac{π}{4})) + 3

, find

y

when

x = \frac{π}{2}

Match each function to its period.

A. $\frac{π}{2}$
B. $π$
C. $2 π$
D. $4 π$

y = sin x

y = tan x

y = sin (4 x)

y = cos (\frac{x}{2})

Why does

y = tan x

have vertical asymptotes at

x = \frac{π}{2} + nπ

Find the maximum value of

y = 4 sin x - 1

A sinusoid has a maximum of

11

, a minimum of

3

, and a period of

π

. Find the value of

k

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