Practice Maths

Transformed Trigonometric Functions and Equations

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Key Terms

The general form is y = a sin(b(x − h)) + k (same for cosine).
Amplitude
= |a| — half the distance from max to min. Negative a reflects in the x-axis.
Period
= 2π/b (for b > 0) — how long before the pattern repeats.
Phase shift
= h — horizontal translation (right if h>0, left if h<0).
Vertical shift
= k — moves the centre line up/down.
For tangent: y = a tan(b(x − h)) + k, period = π/b.
To solve trig equations: find the reference angle, use the unit circle to identify all solutions in the given domain.
📚 QCAA Formula Sheet
Period of sin/cos: T = 2π/b
Period of tan: T = π/b
Amplitude: |a|
General form: y = a sin(b(x − h)) + k
Parameter Effect Example
a = 2Amplitude doublesy=2sin(x): max=2, min=−2
b = 2Period halvesy=sin(2x): period=π
h = π/3Shift right π/3y=sin(x−π/3)
k = 1Shift up 1Centre line y=1
a = −1Reflected in x-axisy=−sin(x): starts at 0 going down
0 π/2 π 3π/2 2π 0 2 −2 y = sin(x) y = 2sin(2x)+1
Hot Tip When solving sin(x) = c over a domain, find the reference angle α = arcsin(|c|). Then apply the CAST rule: sin is positive in Q1 and Q2, negative in Q3 and Q4. Write all solutions as x = α, π−α (for positive) or x = π+α, 2π−α (for negative), then add 2πk for extra periods.

Worked Example 1 — Identifying parameters

Question: For y = 3sin(2x − π/2) + 1, state the amplitude, period, phase shift and vertical shift.

Rewrite in standard form: y = 3sin(2(x − π/4)) + 1

a = 3: Amplitude = 3 (max = 4, min = −2)

b = 2: Period = 2π/2 = π

h = π/4: Phase shift = π/4 right

k = 1: Vertical shift = 1 up, centre line y = 1

Worked Example 2 — Solving a trig equation

Question: Solve 2sin(x) − 1 = 0 for x ∈ [0, 2π].

Rearrange: sin(x) = 1/2

Reference angle: arcsin(1/2) = π/6

sin is positive in Q1 and Q2: x = π/6   or   x = π − π/6 = 5π/6

Solutions: x = π/6 and x = 5π/6

Introduction

Real-world periodic phenomena — tides, sound waves, seasons, heartbeats — are all modelled with transformed trigonometric functions. By adjusting four parameters (a, b, h, k) in y = a sin(b(x − h)) + k, we can match any periodic pattern. This lesson shows you how to read, sketch, and solve equations involving these functions.

The Four Transformations

Start from y = sin(x): period 2π, amplitude 1, passes through the origin. Each parameter changes one feature:

  • |a| stretches the graph vertically — the amplitude. If a < 0, the graph is reflected (starts going down).
  • b compresses the graph horizontally — larger b means shorter period. Period = 2π/b.
  • h slides the graph left or right — the phase shift. Positive h moves it right.
  • k slides the graph up or down — the vertical shift. The new centre line (midline) is y = k.
Worked Example 1 — Reading parameters

Identify all features of y = −2cos(3x + π) + 4.

Step 1: Factor b from the argument: y = −2cos(3(x + π/3)) + 4

Step 2: a = −2 → amplitude = 2, reflected. b = 3 → period = 2π/3. h = −π/3 → phase shift π/3 left. k = 4 → midline y = 4.

Answer: Amplitude 2, period 2π/3, shift π/3 left, midline y = 4, reflected. Max = 6, min = 2.

Sketching Transformed Functions

Use this 5-step method: (1) Find the midline y = k. (2) Mark max = k + |a| and min = k − |a|. (3) Find one complete period starting from the phase shift h. (4) Divide the period into quarters to locate key points. (5) Sketch one full cycle, then extend if needed.

Solving Trigonometric Equations

To solve a sin(bx + c) = d over a given domain:

  1. Isolate the trig function: sin(bx + c) = d/a
  2. Let u = bx + c. Find the domain for u.
  3. Solve sin(u) = d/a for u using the unit circle (all solutions in u-domain).
  4. Back-substitute: x = (u − c)/b
Worked Example 2 — Solving a transformed equation

Solve 2cos(2x) = −√3 for x ∈ [0, 2π].

Step 1: cos(2x) = −√3/2

Step 2: Let u = 2x. As x ∈ [0, 2π], u ∈ [0, 4π].

Step 3: Reference angle: arccos(√3/2) = π/6. cos is negative in Q2 and Q3. u = 5π/6, 7π/6, 5π/6+2π = 17π/6, 7π/6+2π = 19π/6.

Step 4: x = u/2 = 5π/12, 7π/12, 17π/12, 19π/12

💡 Key Reminder: Always factor b out of the argument before identifying the phase shift: y = sin(2x − π) = sin(2(x − π/2)), so the phase shift is π/2 right, not π right. Factoring first prevents this common error.

Summary

y = a sin(b(x − h)) + k has amplitude |a|, period 2π/b, phase shift h, midline y = k. To solve trig equations: isolate the trig function, expand the domain by the b-factor, find all unit-circle solutions, then divide back by b.

Mastery Practice

See Answers ➔
  1. Fluency

    State the amplitude, period, phase shift and vertical shift for each function.

    1. (a) y = 4sin(x)
    2. (b) y = cos(3x)
    3. (c) y = 2sin(x − π/4) + 3
    4. (d) y = −3cos(2x + π/2) − 1
  2. Fluency

    Find the maximum value, minimum value and midline for each function.

    1. (a) y = 5sin(x) − 2
    2. (b) y = −3cos(x) + 4
    3. (c) y = 2sin(x + 1) + 1
    4. (d) y = 4cos(2x) − 3
  3. Fluency

    Solve each equation for x ∈ [0, 2π]. Give exact answers.

    1. (a) sin(x) = 1/2
    2. (b) cos(x) = −1/2
    3. (c) tan(x) = 1
    4. (d) sin(x) = −√3/2
  4. Fluency

    Write the equation of each transformation of y = sin(x).

    1. (a) Amplitude 3, period 2π.
    2. (b) Amplitude 1, period π.
    3. (c) Amplitude 2, period 4π, shift up 5.
    4. (d) Amplitude 1, period 2π, reflected, shift right π/2.
  5. Understanding

    Solve each equation for x ∈ [0, 2π].

    Method: Isolate the trig function, find the reference angle, use CAST to find all solutions in the domain.
    1. (a) 2sin(x) − √3 = 0
    2. (b) 2cos(x) + 1 = 0
    3. (c) √3 tan(x) = 1
    4. (d) 4sin(x) + 2 = 0
  6. Understanding

    Solve each equation over the given domain.

    1. (a) sin(2x) = 1/2 for x ∈ [0, 2π]
    2. (b) cos(2x) = −√2/2 for x ∈ [0, π]
    3. (c) 2sin(x/2) = 1 for x ∈ [0, 4π]
  7. Understanding

    Sketch one complete cycle of each function. Label key points (max, min, x-intercepts, midline).

    1. (a) y = 2sin(x) − 1
    2. (b) y = 3cos(2x) + 1
  8. Understanding

    Find the equation from the graph description.

    1. (a) A sine function with amplitude 4, period π, midline y = 2, no phase shift.
    2. (b) A cosine function with maximum 5, minimum −1, period 4π.
    3. (c) A sine function with amplitude 3, which passes through the origin going downward, and has period 2π.
  9. Problem Solving

    Tide modelling.

    Challenge. The height of the tide (in metres) at a coastal town is modelled by H = 3sin(πt/6) + 5, where t is time in hours after midnight.
    1. (a) State the amplitude, period and midline. Interpret each in context.
    2. (b) What is the maximum tidal height and when does it first occur?
    3. (c) Find all times in the first 24 hours when the tide height is exactly 6.5 m.
  10. Problem Solving

    Matching a function to data.

    Challenge. A Ferris wheel has a diameter of 20 m. The lowest point of a gondola is 2 m above the ground. The wheel completes one full revolution every 40 seconds.
    1. (a) Write an equation h = a cos(bt) + k for the height h (metres) of a gondola at time t (seconds), assuming the gondola starts at the lowest point.
    2. (b) Find the height after 10 seconds and after 25 seconds.
    3. (c) Find all times in the first 40 seconds when the gondola is at a height of 17 m.