Mathematics Advanced • Year 11 • Module 2 • Lesson 10

Graphs of Sine and Cosine

Build fluency in reading amplitude, period, vertical shift, and range from $y = a \sin(bx) + d$ and $y = a \cos(bx) + d$, and in sketching key features.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 For $y = a \sin(bx) + d$, fill in the formulas:

Amplitude = ____________     Period (in radians) = ____________     Range = [____________, ____________]

Q1.2 Five key $x$-values of one cycle of $y = \sin x$ on $[0, 2\pi]$ are 0, π/2, π, 3π/2, 2π. State the $y$-value at each.

x = 0: y = ____   x = π/2: y = ____   x = π: y = ____   x = 3π/2: y = ____   x = 2π: y = ____

Q1.3 Why is the amplitude of $y = -3 \sin x$ equal to 3, not $-3$? Answer in one sentence.

Stuck? Revisit lesson § Two moves — general form at a glance and § Transformations of sine and cosine.

2. Worked example — sketching $y = 2 \sin x$

Every step annotated. Use as the template for the faded version.

Problem. Sketch one cycle of $y = 2 \sin x$ for $0 \leq x \leq 2\pi$ and state its amplitude and range.

Step 1 — Identify the parameters.

a = 2, b = 1, d = 0.

Reason: read directly from the form $y = a \sin(bx) + d$.

Step 2 — Compute amplitude and period.

Amplitude = |a| = 2. Period = 2π/|b| = 2π/1 = 2π.

Reason: amplitude is always positive (Trap 2); period unchanged because $b = 1$.

Step 3 — Plot the five key points.

(0, 0), (π/2, 2), (π, 0), (3π/2, −2), (2π, 0).

Reason: same x-pattern as $\sin x$, but y multiplied by 2.

Step 4 — State the range and conclude.

Range = [−2, 2]; one cycle complete from 0 to 2π.

Conclusion. Amplitude = $\mathbf{2}$, Period = $\mathbf{2\pi}$, Range = $\mathbf{[-2, 2]}$.

3. Faded example — fill in the missing steps

For $y = 3 \cos(2x) + 1$, find amplitude, period, and range. Fill in each blank line. 4 marks

Step 1 — Read off parameters.

a = ____, b = ____, d = ____

Step 2 — Amplitude.

Amplitude = |a| = ________

Step 3 — Period.

Period = 2π / |b| = 2π / ____ = ________

Step 4 — Range.

Range = [d − |a|, d + |a|] = [____, ____]

Conclusion. Amplitude = ____, Period = ____, Range = ____.

Stuck? Revisit lesson § Worked Example — Graph with vertical shift.

4. Graduated practice

State amplitude, period, and (where there is a vertical shift) range. For one question, sketch.

Foundation — basic transforms (4 questions)

QFunctionAmplitudePeriod
4.1 1$y = \sin x$
4.2 1$y = 2 \cos x$
4.3 1$y = \sin(2x)$
4.4 1$y = \cos\left(\frac{x}{2}\right)$

Standard — transformations + ranges (6 questions)

4.5 Find the amplitude, period, and range of $y = 4 \sin(3x)$.    2 marks

4.6 Find the amplitude, period, and range of $y = \cos x + 2$.    2 marks

4.7 Find the amplitude, period, and range of $y = -2 \cos(4x) + 3$. (Mind the negative amplitude coefficient.)    2 marks

4.8 Find the amplitude, period, and range of $y = 5 \sin\left(\frac{x}{3}\right) - 1$.    2 marks

4.9 A student claims the period of $y = \sin(2x)$ is $4\pi$. Identify which trap from the lesson this commits and give the correct period.    2 marks

4.10 Sketch one cycle of $y = 3 \sin x$ for $0 \leq x \leq 2\pi$. Label the five key points $(x, y)$, the amplitude, and the period.    3 marks

Extension — combine ideas (2 questions)

4.11 Find amplitude, period, and range of $y = -3 \cos\left(\frac{\pi x}{2}\right) + 4$. State the period in the same units as $x$.    3 marks

4.12 The graph of $y = a \sin(bx) + d$ has maximum value 7, minimum value $-3$, and period $\pi$. Find $a$ (positive), $b$ (positive), and $d$. Show every step.    3 marks

Stuck on 4.12? $d$ = midline = $\frac{\max + \min}{2}$; $|a|$ = amplitude = $\frac{\max - \min}{2}$; $b = \frac{2\pi}{\text{period}}$.

5. Self-check the easy 3

Tick the first three once you've verified your method works.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Formulas

Amplitude = $|a|$.   Period = $\frac{2\pi}{|b|}$.   Range = $[d - |a|, d + |a|]$.

Q1.2 — Five key points of $y = \sin x$

$x = 0$: $y = 0$.   $x = \pi/2$: $y = 1$.   $x = \pi$: $y = 0$.   $x = 3\pi/2$: $y = -1$.   $x = 2\pi$: $y = 0$.

Q1.3 — Why amplitude is positive

Amplitude is a distance from the midline to the peak; distance is always positive. The negative sign on $a = -3$ instead reflects the graph in the $x$-axis (Trap 2 in the lesson).

Q3 — Faded example: $3 \cos(2x) + 1$

Step 1: $a = 3$, $b = 2$, $d = 1$.
Step 2: Amplitude $= |3| = 3$.
Step 3: Period $= 2\pi / 2 = \pi$.
Step 4: Range $= [1 - 3, 1 + 3] = [-2, 4]$.
Conclusion: Amplitude $= \mathbf{3}$, Period $= \mathbf{\pi}$, Range $= \mathbf{[-2, 4]}$.

Q4.1 — $\sin x$

Amplitude $= 1$, Period $= 2\pi$.

Q4.2 — $2 \cos x$

Amplitude $= 2$, Period $= 2\pi$.

Q4.3 — $\sin(2x)$

Amplitude $= 1$, Period $= \frac{2\pi}{2} = \pi$.

Q4.4 — $\cos\left(\frac{x}{2}\right)$

$b = \frac{1}{2}$. Amplitude $= 1$, Period $= \frac{2\pi}{1/2} = 4\pi$.

Q4.5 — $4 \sin(3x)$

Amplitude $= 4$, Period $= \frac{2\pi}{3}$, Range $= [-4, 4]$.

Q4.6 — $\cos x + 2$

Amplitude $= 1$, Period $= 2\pi$, Range $= [2 - 1, 2 + 1] = [1, 3]$.

Q4.7 — $-2 \cos(4x) + 3$

Amplitude $= |-2| = 2$, Period $= \frac{2\pi}{4} = \frac{\pi}{2}$, Range $= [3 - 2, 3 + 2] = [1, 5]$.

Q4.8 — $5 \sin\left(\frac{x}{3}\right) - 1$

Amplitude $= 5$, Period $= \frac{2\pi}{1/3} = 6\pi$, Range $= [-1 - 5, -1 + 5] = [-6, 4]$.

Q4.9 — Critique of "period of $\sin(2x)$ is $4\pi$"

This commits Trap 1 in the lesson: $b = 2$ compresses the graph (halves the period) rather than stretching. Correct period: $\frac{2\pi}{2} = \mathbf{\pi}$.

Q4.10 — Sketch $y = 3 \sin x$

Amplitude = 3, period = $2\pi$. Five key points: $(0, 0)$, $\left(\frac{\pi}{2}, 3\right)$, $(\pi, 0)$, $\left(\frac{3\pi}{2}, -3\right)$, $(2\pi, 0)$. Sketch is a smooth sine wave oscillating between $y = -3$ and $y = 3$ across one cycle. Range $= [-3, 3]$.

Q4.11 — $-3 \cos\left(\frac{\pi x}{2}\right) + 4$

$a = -3$, $b = \frac{\pi}{2}$, $d = 4$. Amplitude $= |-3| = 3$. Period $= \frac{2\pi}{\pi/2} = \frac{2\pi \cdot 2}{\pi} = 4$ (in the same units as $x$). Range $= [4 - 3, 4 + 3] = [1, 7]$.

Q4.12 — Find $a$, $b$, $d$

$d = \frac{\max + \min}{2} = \frac{7 + (-3)}{2} = \mathbf{2}$.
Amplitude $= \frac{\max - \min}{2} = \frac{7 - (-3)}{2} = 5$, so $a = \mathbf{5}$ (taking positive).
$b = \frac{2\pi}{\text{period}} = \frac{2\pi}{\pi} = \mathbf{2}$.
So $y = 5 \sin(2x) + 2$. Check: max $= 5 + 2 = 7$ ✓, min $= -5 + 2 = -3$ ✓, period $= \frac{2\pi}{2} = \pi$ ✓.