The parent curves $y = \sin x$, $y = \cos x$ and $y = \tan x$ can be stretched, flipped, and shifted in any direction. Mastering the general form $y = a \cdot f(bx + c) + d$ lets you read off amplitude, period, phase shift and vertical translation at a glance, and sketch or solve any transformed trig graph.
Today's hook, A Ferris wheel with radius 15 m has its centre 20 m above the ground. It completes one full rotation every 40 seconds. Which function models the height of a passenger above the ground? And what transformations distinguish it from a plain $\sin$ curve?
0/5QUESTS
1
Orient to the transformation toolkit
Predict the Ferris wheel model, set your goals and recall the language of transformations.
01
Recall, your gut answer first
+5 XP warm-up
A Ferris wheel with radius 15 m has its centre 20 m above the ground. It completes one rotation every 40 seconds. Write your gut answers before the lesson, no calculating yet:
What is the highest and lowest point a passenger reaches?
How long does one complete up-and-down cycle take?
What transformations would change a plain $\sin x$ graph into the Ferris wheel model?
auto-saved
02
What you'll master
Know
Key facts
The general form $y = a \cdot f(bx + c) + d$ and what each parameter controls
Amplitude $= |a|$; period $= \dfrac{2\pi}{|b|}$ for sin/cos; $\dfrac{\pi}{|b|}$ for tan
Phase shift $= -\dfrac{c}{b}$; midline $= d$
Understand
Concepts
How reflections, dilations and translations act independently on a trig graph
Why a negative $a$ reflects the graph but doesn't change the amplitude value
How phase shift and period together determine all solution positions in a domain
Can do
Skills
Identify all transformation parameters from a given equation
Sketch a transformed trig graph by finding 5 key points over one period
Solve trig equations graphically within a specified domain
03
Key terms
AmplitudeHalf the total height of the wave: $|a|$. Distance from the midline to a peak or trough.
PeriodThe horizontal length of one complete cycle. For sin/cos: $\dfrac{2\pi}{|b|}$. For tan: $\dfrac{\pi}{|b|}$.
Phase shiftHorizontal translation of the graph. Given by $-\dfrac{c}{b}$. Positive = shift right; negative = shift left.
Vertical translationThe value $d$ shifts the entire graph up ($d > 0$) or down ($d < 0$). Also called the vertical shift.
DilationStretching or compressing. $|a| \neq 1$ dilates vertically; $|b| \neq 1$ dilates horizontally (changing the period).
MidlineThe horizontal line $y = d$ that the wave oscillates around. Halfway between the maximum and minimum.
Domain restrictionThe specified interval for $x$ (e.g. $0 \le x \le 2\pi$) that limits how many solution cycles exist.
2
Apply the transformation toolkit
Connect each parameter in the general form to its effect on a parent trigonometric graph.
04
The Toolkit: Reflections, Translations and Dilations
core concept
Every transformation of a trig function is a modification to one of four parameters in the general form. Recognising which parameter does what is the master key to this entire topic.
Starting from the base curves $y = \sin x$, $y = \cos x$, $y = \tan x$:
Vertical reflection: $y = -\sin x$ flips the graph over the $x$-axis (peaks become troughs).
Horizontal reflection: $y = \sin(-x) = -\sin x$ also flips vertically for sine (since sine is odd). For cosine, $y = \cos(-x) = \cos x$, no change (cosine is even).
Vertical translation: $y = \sin x + d$ shifts the graph up or down by $d$ units.
Horizontal translation (phase shift): $y = \sin(x + c)$ shifts left by $c$ units (for $c > 0$).
Vertical dilation: $y = a \cdot \sin x$ changes the amplitude to $|a|$.
Horizontal dilation: $y = \sin(bx)$ changes the period to $\dfrac{2\pi}{|b|}$.
Master formula: All four transformations combine into one expression. Order of operations: first apply the horizontal dilation and phase shift inside the function, then the vertical dilation outside, then the vertical translation.
$$y = a \cdot f(bx + c) + d$$
$a$ controls the amplitude
$|a|$ is the amplitude. If $a < 0$, the graph is also reflected over the $x$-axis. A negative $a$ does NOT make the amplitude negative, amplitude is always positive.
$b$ controls the period
Larger $|b|$ = faster oscillation = shorter period. $b = 2$ halves the period; $b = \frac{1}{2}$ doubles it. Period $= \dfrac{2\pi}{|b|}$ for sin/cos.
$c$ shifts left/right
Watch the sign: $y = \sin(x + \frac{\pi}{4})$ shifts LEFT by $\frac{\pi}{4}$. The phase shift is $-\frac{c}{b}$, so a positive $c$ means a leftward shift.
$y = a \cdot f(bx + c) + d$: $a$ = amplitude/reflection, $b$ = period change, $c$ = phase shift, $d$ = vertical shift.; Vertical reflection: $a < 0$ flips the graph. Amplitude = $|a|$ (always positive).
Pause, copy the general form $y = a \cdot f(bx+c)+d$ with the role of each parameter: $a$ = amplitude and reflection, $b$ = period change (period = $2\pi/|b|$), $c/b$ = phase shift, $d$ = vertical shift, into your book.
Quick check: For $y = -4\sin(3x + \pi) - 2$, which statement about the parameter $a$ is correct?
3
Determine amplitude and period
Read vertical and horizontal dilations from transformed sine, cosine and tangent functions.
05
Amplitude and Period
core concept
We just saw the four parameters of $y = a \cdot f(bx+c)+d$: $a$, $b$, $c$, and $d$ each change the shape or position. That raises a question: for $a$ and $b$ specifically, what is the exact formula for amplitude and period, and why is amplitude always positive even when $a < 0$? This card answers it → amplitude $= |a|$ (max $= d + |a|$, min $= d - |a|$); period $= 2\pi/|b|$ for sin/cos, $\pi/|b|$ for tan.
Amplitude and period are the two most commonly tested features of a transformed trig graph. Amplitude tells you how tall the wave is; period tells you how long one cycle takes.
Definitions and formulas:
Amplitude $= |a|$: The distance from the midline to either a peak or a trough. Maximum $= d + |a|$; minimum $= d - |a|$.
Period of sin/cos $= \dfrac{2\pi}{|b|}$: How far along the $x$-axis before the pattern repeats.
Period of tan $= \dfrac{\pi}{|b|}$: Tan has a shorter natural period of $\pi$, not $2\pi$.
Midline $= y = d$: The equilibrium line. Even if $d = 0$, the midline is the $x$-axis.
Worked example, $y = 3\sin(2x)$:
Amplitude $= |3| = 3$
Period $= \dfrac{2\pi}{|2|} = \pi$
Midline: $y = 0$ (no vertical shift)
Maximum value: $0 + 3 = 3$. Minimum value: $0 - 3 = -3$.
The graph completes one full cycle from $x = 0$ to $x = \pi$.
These two formulas are the most frequently examined. Memorise them exactly, the period formula is especially prone to errors when students forget the absolute value around $b$.
Why $|b|$ not $b$? A negative $b$ reflects horizontally. $y = \sin(-2x) = -\sin(2x)$, so the period is still $\dfrac{2\pi}{2} = \pi$, the same as for $b = 2$. Always use the absolute value when computing the period.
Amplitude $= |a|$. Max $= d + |a|$, min $= d - |a|$. Never negative.; Period (sin/cos) $= 2\pi / |b|$. Period (tan) $= \pi / |b|$.
Pause, copy amplitude $= |a|$, max $= d + |a|$, min $= d - |a|$, period $= 2\pi/|b|$ (sin/cos) and $\pi/|b|$ (tan), noting amplitude is always positive, even when $a < 0$, into your book.
True or false: The amplitude of $y = -5\cos(3x) + 2$ is $-5$.
False. Amplitude $= |a| = |-5| = 5$. Amplitude is always positive. The $-5$ tells you the graph is reflected over the $x$-axis, but the amplitude is still 5.
4
Determine phase and vertical shifts
Factor the input correctly, locate the midline and interpret the direction of each translation.
06
Phase Shift and Vertical Shift
core concept
We just saw that amplitude $= |a|$ and period $= 2\pi/|b|$ describe the shape of the wave. That raises a question: the parameters $c$ and $d$ shift the entire graph horizontally and vertically, but why is the phase shift $-c/b$, not simply $c$, and why must you factor out $b$ first? This card answers it → rewrite as $y = a\sin\!\left(b\!\left(x + c/b\right)\!\right) + d$; the phase shift is $-c/b$ (positive = right, negative = left).
Phase shift moves the entire graph left or right along the $x$-axis. Vertical shift moves it up or down. Both leave the shape (amplitude and period) unchanged.
Reading phase shift carefully:
Phase shift $= -\dfrac{c}{b}$: In $y = a \cdot f(bx + c) + d$, the phase shift is $-c/b$.
Positive result → shift to the right. Negative result → shift to the left.
In $y = \sin(x + c)$ (where $b = 1$): phase shift $= -c$. So $y = \sin(x + \pi/4)$ shifts left by $\pi/4$.
Vertical shift $= d$: Positive $d$ raises the midline; negative $d$ lowers it.
Full worked example, $y = 2\cos\!\left(x - \dfrac{\pi}{3}\right) + 1$:
$a = 2$: amplitude $= 2$, no reflection.
$b = 1$: period $= \dfrac{2\pi}{1} = 2\pi$ (unchanged from base cosine).
$c = -\dfrac{\pi}{3}$: phase shift $= -\dfrac{c}{b} = -\dfrac{-\pi/3}{1} = +\dfrac{\pi}{3}$ (shift right $\dfrac{\pi}{3}$).
$d = 1$: midline $y = 1$. Max $= 1 + 2 = 3$; min $= 1 - 2 = -1$.
Watch the bracket sign
$\sin(x + \pi/4)$ shifts LEFT. $\sin(x - \pi/4)$ shifts RIGHT. The direction is opposite to the sign inside the bracket. Many students get this backwards.
Factor out $b$ first
For $y = \sin(2x + \pi)$, factor: $y = \sin\!\left(2\!\left(x + \dfrac{\pi}{2}\right)\!\right)$. Phase shift $= -\dfrac{\pi}{2}$ (left $\dfrac{\pi}{2}$). Always factor out $b$ before reading phase shift.
Phase shift $= -c/b$. Positive → right; negative → left.; To read phase shift from $y = a\sin(bx + c) + d$: factor out $b$ first: $y = a\sin\!\left(b\!\left(x + c/b\right)\!\right) + d$. Phase shift $= -c/b$.
Pause, copy the phase-shift rule: factor out $b$ first from $y = a\sin(bx+c)+d$, giving phase shift $= -c/b$ (positive = right, negative = left); vertical shift $= d$ (midline $y = d$), into your book.
Fill the blanks for $y = 3\sin\!\left(2x - \dfrac{\pi}{2}\right) + 4$: Factor out $b$: $y = 3\sin\!\left(2\!\left(x - \dfrac{\pi}{4}\right)\!\right) + 4$. Amplitude $= $ . Period $= $ (type "pi"). Phase shift $= $ right (type "pi/4"). Midline: $y = $ .
5
Sketch transformed trigonometric graphs
Use five key points to build an accurate cycle with labelled features.
07
Sketching Transformed Trig Graphs
core concept
We just saw all four parameters: amplitude, period, phase shift $-c/b$, and vertical shift $d$. That raises a question: with all these values in hand, what is the minimum number of points you need to plot to produce an accurate sketch, and where exactly should those points go? This card answers it → the five-key-point method: start at the phase shift, then step through $T/4$, $T/2$, $3T/4$, and $T$ to capture one complete cycle.
To sketch a transformed trig graph, you don't need to plot dozens of points. The five-key-point method reduces any transformed sine or cosine to exactly five strategic points that define one complete cycle.
Five-key-point method, step by step:
Step 1: Identify $a$, $b$, $c$, $d$ from the equation.
Step 2: Calculate amplitude, period, phase shift and midline.
Step 3: Divide one period into 4 equal quarters to find the $x$-values of the 5 key points.
Step 4: For each key $x$, calculate the $y$-value using the pattern: start → max/min → midline → min/max → midline (for sin); or max → midline → min → midline → max (for cos).
Step 5: Apply any vertical shift $d$ to all $y$-values. Plot and join with a smooth curve.
Worked example, $y = -2\sin(\pi x + \pi)$:
Factor: $y = -2\sin(\pi(x + 1))$
$a = -2$: amplitude $= 2$, reflected. $b = \pi$: period $= \dfrac{2\pi}{\pi} = 2$. Phase shift $= -1$ (left 1). $d = 0$: midline $y = 0$.
Key $x$-values (one period starting at phase shift $x = -1$): $-1,\ -\tfrac{1}{2},\ 0,\ \tfrac{1}{2},\ 1$.
Without reflection: $0 \to 2 \to 0 \to -2 \to 0$. With $a = -2$ (reflected): $0 \to -2 \to 0 \to 2 \to 0$.
The graph starts at the midline, dips to $-2$ (quarter period), returns to midline (half period), rises to $+2$ (three-quarter period), and returns to midline (full period).
HSC tip, labelling the sketch: Always label (1) all $x$-intercepts, (2) coordinates of every maximum and minimum, (3) the midline $y = d$, (4) the period on the $x$-axis. Missing labels cost marks even when the shape is correct.
Five-key-point method: phase shift → phase shift + T/4 → phase shift + T/2 → phase shift + 3T/4 → phase shift + T.; For reflected graphs ($a < 0$): flip all $y$-values. A sine that would peak first now troughs first.
Pause, copy the five-key-point $x$-values: phase shift; $+T/4$; $+T/2$; $+3T/4$; $+T$, and the rule that when $a < 0$, flip all $y$-values so a sine starts with a trough, into your book.
Match each equation feature to its correct description:
Amplitude = |a|
Period = 2π/|b|
Phase shift = −c/b
Midline y = d
Horizontal distance the graph is shifted left or right
The equilibrium line the wave oscillates around
Distance from midline to peak or trough
Horizontal length of one complete sine/cosine cycle
6
Solve, check and consolidate
Solve graphically, test your understanding and revisit the opening Ferris wheel model.
08
Solving Trig Equations Graphically
core concept
We just saw the five-key-point method for sketching one complete cycle of a transformed trig graph. That raises a question: if you need to find all $x$-values where a transformed trig function equals a constant $k$, how do you use the sketch, and the substitution $u = bx + c$, to find every solution in the domain? This card answers it → isolate the trig ratio, substitute $u = bx + c$, adjust the domain for $u$, find all $u$-solutions using symmetry, then back-substitute for $x$.
A trig equation within a domain is asking: for what $x$-values does the transformed graph intersect a horizontal line? Finding all solutions requires understanding the period and domain together.
Graphical approach to solving trig equations:
Set up: Rewrite as $f(\text{transformed}) = k$. Sketch the transformed graph over the given domain.
Draw $y = k$: The solutions are the $x$-coordinates of all intersection points within the domain.
Count solutions: Use the period, each full period of the graph can produce at most 2 solutions for sin/cos (one on the way up, one on the way down).
Domain restriction: Only include solutions that fall within the given interval. Clearly justify why no others exist.
Worked example, solve $2\sin\!\left(x - \dfrac{\pi}{6}\right) = 1$ for $0 \le x \le 2\pi$:
Isolate: $\sin\!\left(x - \dfrac{\pi}{6}\right) = \dfrac{1}{2}$
Let $u = x - \dfrac{\pi}{6}$. Range of $u$: when $x \in [0, 2\pi]$, $u \in \left[-\dfrac{\pi}{6},\ \dfrac{11\pi}{6}\right]$.
$\sin u = \dfrac{1}{2}$ → reference angle $\dfrac{\pi}{6}$. In $\left[-\dfrac{\pi}{6},\ \dfrac{11\pi}{6}\right]$: $u = \dfrac{\pi}{6}$ (1st quadrant) and $u = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ (2nd quadrant).
Back-substitute: $x = u + \dfrac{\pi}{6}$:
$x = \dfrac{\pi}{6} + \dfrac{\pi}{6} = \dfrac{\pi}{3}$ and $x = \dfrac{5\pi}{6} + \dfrac{\pi}{6} = \pi$.
Both lie in $[0, 2\pi]$. Period $= 2\pi$, so no additional cycles exist in this domain. Two solutions: $x = \dfrac{\pi}{3}$ and $x = \pi$.
To solve $a\sin(bx + c) + d = k$: isolate the trig function, substitute $u = bx + c$, adjust the domain for $u$, find all $u$-solutions, back-substitute for $x$.; For each full period in the domain, sin/cos can produce 2 solutions...
Pause, copy the solving method: isolate the trig ratio → substitute $u = bx + c$ → adjust domain for $u$ → find all $u$-solutions using reference angle and symmetry → back-substitute to find $x$, into your book.
Top 3 list: List THREE real-world contexts where a transformed trig function (not just a plain $\sin x$) would be the best model. For each, name at least one transformation that distinguishes it from the plain curve.
Worked examples · 3 problems
PROBLEM 1 · READING PARAMETERS
For $y = -3\cos\!\left(2x - \dfrac{\pi}{2}\right) - 1$, state: (a) amplitude, (b) period, (c) phase shift, (d) midline, (e) maximum and minimum values.
1
Identify parameters $a = -3,\quad b = 2,\quad c = -\dfrac{\pi}{2},\quad d = -1$
Compare with $y = a\cos(bx + c) + d$. The coefficient of $x$ inside the bracket is $b = 2$, and the constant inside is $c = -\pi/2$.
2
Parts (a) and (b) Amplitude $= |a| = |-3| = 3$ Period $= \dfrac{2\pi}{|b|} = \dfrac{2\pi}{2} = \pi$
Always take the absolute value of $a$ for amplitude. The negative sign means the graph is reflected, not that the amplitude is $-3$.
3
Parts (c), (d), (e) Factor: $y = -3\cos\!\left(2\!\left(x - \dfrac{\pi}{4}\right)\!\right) - 1$ Phase shift $= +\dfrac{\pi}{4}$ (right) Midline: $y = -1$ Max $= d + |a| = -1 + 3 = 2$ Min $= d - |a| = -1 - 3 = -4$
Factor out $b = 2$ to expose the phase shift directly: $c/b = (-\pi/2)/2 = -\pi/4$, so phase shift $= +\pi/4$ (rightward). Since $a = -3$ (negative), the usual cosine peak becomes a trough and vice versa.
PROBLEM 2 · SKETCHING WITH FIVE KEY POINTS
Sketch $y = 2\sin\!\left(\dfrac{x}{2} + \dfrac{\pi}{4}\right)$ for $-\pi \le x \le 3\pi$. Label all intercepts, maxima and minima.
Period = $4\pi$ means one full cycle spans $4\pi$ units. With domain $[-\pi, 3\pi]$ (width $4\pi$), there is exactly one full cycle to sketch.
2
Five key $x$-values (start at phase shift $-\pi/2$, step by T/4 $= \pi$): $x = -\dfrac{\pi}{2},\quad \dfrac{\pi}{2},\quad \dfrac{3\pi}{2},\quad \dfrac{5\pi}{2},\quad \dfrac{7\pi}{2}$ But domain starts at $x = -\pi$, so only $x \in [-\pi, 3\pi]$. Trim to domain: use $x = -\pi/2$ to $x = 7\pi/2$ but note $7\pi/2 > 3\pi$, so mark end at $3\pi$.
The key points define the shape. Points outside the domain are not plotted, but you need to know where the cycle sits to identify which portion falls in $[-\pi, 3\pi]$.
3
$y$-values at key $x$-values: $(-\pi/2,\ 0)$: start of cycle (midline) $(\pi/2,\ 2)$: maximum $(3\pi/2,\ 0)$: midline crossing $(5\pi/2,\ -2)$: minimum Sketch also extends left to $x = -\pi$: $y = 2\sin(\frac{-\pi}{2} + \frac{\pi}{4}) = 2\sin(-\frac{\pi}{4}) = -\sqrt{2}$. Label: max at $(\pi/2, 2)$; min at $(5\pi/2, -2)$; intercepts at $x = -\pi/2,\ 3\pi/2$ (and boundary at $x = 3\pi$).
Plot only points in the domain. At the left boundary $x = -\pi$, calculate the exact $y$-value rather than assuming it starts at zero. Always check boundary values.
PROBLEM 3 · SOLVING A TRIG EQUATION GRAPHICALLY
Solve $2\cos\!\left(x - \dfrac{\pi}{4}\right) = \sqrt{2}$ for $0 \le x \le 2\pi$. Show all working and justify the number of solutions.
1
Isolate the trig function $\cos\!\left(x - \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$ Let $u = x - \dfrac{\pi}{4}$ Domain of $u$: $x \in [0, 2\pi]$ so $u \in \left[-\dfrac{\pi}{4},\ \dfrac{7\pi}{4}\right]$
Dividing both sides by 2 gives the standard form $\cos u = \sqrt{2}/2$. Adjusting the domain for $u$ ensures we find ALL solutions in the original domain.
2
Solve $\cos u = \dfrac{\sqrt{2}}{2}$ in $\left[-\dfrac{\pi}{4},\ \dfrac{7\pi}{4}\right]$ Reference angle: $\cos^{-1}\!\left(\dfrac{\sqrt{2}}{2}\right) = \dfrac{\pi}{4}$ Cosine is positive in Q1 and Q4, and the $u$ domain starts BELOW zero, so sweep the whole interval: $u = -\dfrac{\pi}{4}$ ✓ (in domain, at the lower boundary) $u = \dfrac{\pi}{4}$ (Q1) ✓ (in domain) $u = 2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$ (Q4) ✓ (in domain, at the upper boundary)
$\sqrt{2}/2 \approx 0.707$ is the exact cosine of $\pi/4$, and cosine is even, so $\cos\!\left(-\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$ as well. Shifting the domain moved its lower end to $-\dfrac{\pi}{4}$, so the negative solution is inside the interval and must be counted. Listing only the Q1 and Q4 angles between $0$ and $2\pi$ is what loses it.
3
Back-substitute $x = u + \dfrac{\pi}{4}$: $x = -\dfrac{\pi}{4} + \dfrac{\pi}{4} = 0$ $x = \dfrac{\pi}{4} + \dfrac{\pi}{4} = \dfrac{\pi}{2}$ $x = \dfrac{7\pi}{4} + \dfrac{\pi}{4} = 2\pi$ All three in $[0, 2\pi]$. ✓ $\therefore x = 0,\ \dfrac{\pi}{2},\ 2\pi$
Justification: over an interval of exactly one period, $\cos u = \sqrt{2}/2$ is met twice in the OPEN interval, but this domain is CLOSED at both ends and both endpoints satisfy the equation. $x = 0$ and $x = 2\pi$ are the same phase, yet they are two distinct allowed values of $x$, so the count here is three, not two. Check each candidate against the closed domain rather than assuming one period always gives two solutions.
Multiple choice · 5 questions
Q1. What is the amplitude of $y = -3\sin(2x) + 1$?
Amplitude $= |a| = |-3| = 3$. The $+1$ is a vertical shift, not part of the amplitude. The negative sign reflects the graph but does not affect the amplitude value.
Q2. What is the period of $y = \cos\!\left(\dfrac{\pi x}{2}\right)$?
Q3. The graph of $y = \sin\!\left(x + \dfrac{\pi}{4}\right)$ is the graph of $y = \sin x$ shifted:
Phase shift $= -c/b = -(\pi/4)/1 = -\pi/4$. A negative phase shift means the graph moves to the LEFT by $\pi/4$. Remember: positive $c$ inside the bracket means leftward shift.
Q4. What is the equation of the midline of $y = 2\cos(x) - 3$?
The midline is $y = d$. Here $d = -3$, so the midline is $y = -3$. The amplitude of 2 tells you the wave extends to $y = -3 + 2 = -1$ (max) and $y = -3 - 2 = -5$ (min).
Q5. Which equation has amplitude 4 and period $\pi$?
Need $|a| = 4$ and $2\pi/|b| = \pi$, so $|b| = 2$. Only option A gives $a = 4$ and $b = 2$. Option B has period $2\pi$; option C has amplitude 2; option D has period $\pi/2$.
10
Revisit your thinking
Back to the Ferris wheel: radius 15 m, centre 20 m above ground, one rotation in 40 seconds. The height function is $h(t) = 15\sin\!\left(\dfrac{2\pi}{40}t - \dfrac{\pi}{2}\right) + 20 = 15\sin\!\left(\dfrac{\pi}{20}t - \dfrac{\pi}{2}\right) + 20$. Transformations from plain $\sin$: amplitude $= 15$ (vertical dilation by 15), period $= 40$ (horizontal dilation with $b = \pi/20$), vertical shift $+20$ (centre height), phase shift (passenger starts at the bottom at $t = 0$, the $-\pi/2$ shifts the curve right so it starts at a trough).
auto-saved
7
Complete independent practice
Answer the lesson questions, show your working and compare your reasoning with the worked solutions.
01
Short answer, exam-style questions
show all working
ApplyBand 33 marks
SA 1. State the amplitude, period, phase shift and vertical shift of $y = -2\cos\!\left(3x - \dfrac{\pi}{2}\right) + 4$. Hence state the maximum and minimum values of the function, and sketch one complete cycle. (3 marks)
auto-saved
ApplyBand 44 marks
SA 2. Sketch the graph of $y = 3\sin(2x + \pi)$ for $0 \le x \le 2\pi$. Label all $x$-intercepts, maxima and minima with their coordinates. (4 marks)
auto-saved
AnalyseBand 55 marks
SA 3. Solve $2\cos\!\left(x - \dfrac{\pi}{3}\right) = \sqrt{3}$ for $0 \le x \le 2\pi$. Show all working and justify why there are no other solutions in the domain. (5 marks)
auto-saved
📖 Comprehensive answers (click to reveal)
SA 1 (3 marks): Factor: $y = -2\cos\!\left(3\!\left(x - \dfrac{\pi}{6}\right)\!\right) + 4$. Amplitude $= |-2| = 2$ [1]. Period $= \dfrac{2\pi}{3}$ [1]. Phase shift $= +\dfrac{\pi}{6}$ (right $\dfrac{\pi}{6}$). Vertical shift $= +4$, midline $y = 4$. Max $= 4 + 2 = 6$; min $= 4 - 2 = 2$. Since $a = -2$ (negative), the graph starts at a peak (not a trough) at the phase-shifted start, i.e., at $x = \pi/6$, $y = 6$ (maximum). Sketch: one period from $x = \pi/6$ to $x = \pi/6 + 2\pi/3 = 5\pi/6$, starting at the maximum $6$, descending to minimum $2$ at the midpoint, and returning to $6$ [1].
SA 2 (4 marks): $a = 3$, $b = 2$, $c = \pi$, $d = 0$. Factor: $y = 3\sin\!\left(2\!\left(x + \dfrac{\pi}{2}\right)\!\right)$. Amplitude $= 3$ [1]. Period $= \dfrac{2\pi}{2} = \pi$ [1]. Phase shift $= -\dfrac{\pi}{2}$ (left $\dfrac{\pi}{2}$). Two full periods in $[0, 2\pi]$. Key points (period 1, starting from $x = -\pi/2$ but domain starts at $0$): at $x = 0$: $y = 3\sin(\pi) = 0$; at $x = \pi/4$: $y = 3\sin(\pi + \pi/2) = 3\sin(3\pi/2) = -3$ (min); at $x = \pi/2$: $y = 0$ (intercept); at $x = 3\pi/4$: $y = 3$ (max); at $x = \pi$: $y = 0$. Period 2 repeats: min at $(5\pi/4, -3)$, intercept at $(3\pi/2, 0)$, max at $(7\pi/4, 3)$, end at $(2\pi, 0)$ [1 for key points; 1 for correct labelling of all maxima, minima and intercepts].
SA 3 (5 marks): Isolate: $\cos\!\left(x - \dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2}$ [1]. Let $u = x - \dfrac{\pi}{3}$; $x \in [0, 2\pi] \Rightarrow u \in \left[-\dfrac{\pi}{3},\ \dfrac{5\pi}{3}\right]$ [1]. Reference angle: $\cos^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{6}$. Cosine positive in Q1 and Q4: $u = \dfrac{\pi}{6}$ (Q1) ✓; $u = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$, check: $\dfrac{11\pi}{6} > \dfrac{5\pi}{3}$, outside domain. Next Q1 solution at $u = \dfrac{\pi}{6} - 2\pi < -\dfrac{\pi}{3}$, also outside [1]. Back-substitute: $x = u + \dfrac{\pi}{3}$: only $x = \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi}{2}$ [1]. Justification: period $= 2\pi$ and the domain has width $2\pi$, so at most 2 solutions per period. The Q4 solution for $u$ falls just outside the adjusted domain, leaving only one solution in $[0, 2\pi]$: $x = \dfrac{\pi}{2}$ [1].
8
Review and finish
Retrieve this lesson's transformation method, then extend your practice with the module quiz.
R
Retrieve the transformation method
review
Without looking back, explain how $a$, $b$, $c$ and $d$ control amplitude, period, phase shift and midline in $y=a\cdot f(bx+c)+d$. Then list the five key $x$-positions used to sketch one cycle and the substitution steps used to solve a transformed trigonometric equation on a restricted domain.
auto-saved
01
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.