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S5 Maths: Trigonometric Graphs and Transformations: Paper 3 (challenge)
S5 Maths: Trigonometric Graphs and Transformations: Paper 3 (challenge)
Ten challenge items: recovering amplitude and centre from extreme values, rebuilding an equation from extremes and period, and shifts in combined transformations.
Direct answers Items 1-3: maximum 3 and negative 1, 120 degrees, 1. Items 4-6: 4, y = 2, 60 degrees. Items 7-9: maximum 8 and 2, y = 3 + 2 sin 2x, right by 30 degrees. Item 10: the amplitude should be 2.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S5 Maths: Trigonometric Graphs and Transformations |
| Level | Challenge |
02 Questions (10 items)
- FoundationFind the maximum and minimum of y = 2 sin 3x + 1.
- FoundationWhat is the period of y = 2 sin 3x + 1 in degrees?
- FoundationWhat is the maximum value of y = 4 sin 2x – 3?
- IntermediateA curve has maximum 6 and minimum negative 2. Find its amplitude.
- IntermediateA curve has maximum 6 and minimum negative 2. Find its vertical centre line.
- IntermediateWhat is the period of y = tan 3x in degrees?
- ChallengeFind the maximum and minimum of y = 5 – 3 cos x.
- ChallengeA curve has maximum 5, minimum 1 and period 180 degrees, with its maximum at x = 45 degrees. Write its equation using sine.
- ChallengeHow is y = sin(3x – 90 degrees) shifted from y = sin 3x?
- Error redoA student gives amplitude 3 and period 120 degrees for y = 2 sin 3x. Which is right, which is wrong, and what is the correct value?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Maximum 3, minimum negative 1 | The sine part runs from negative 2 to 2, and adding 1 gives 3 and negative 1. |
| 2 | 120 degrees | The multiplier on x is 3, so 360 divided by 3 is 120 degrees. |
| 3 | 1 | Sine peaks at 1, times 4 gives 4, and subtracting 3 gives 1. |
| 4 | 4 | Amplitude is the difference between maximum and minimum halved, so 8 divided by 2 is 4. |
| 5 | y = 2 | The centre line is the sum of maximum and minimum halved, so 4 divided by 2 is 2. |
| 6 | 60 degrees | The basic tangent period is 180 degrees, and dividing by 3 gives 60 degrees. |
| 7 | Maximum 8, minimum 2 | cos runs from negative 1 to 1, times negative 3 gives 3 to negative 3, and adding 5 gives 8 and 2. |
| 8 | y = 3 + 2 sin 2x | Centre 3, amplitude 2 and multiplier 2 on x give this form, and sin 2x reaches 1 at 45 degrees. |
| 9 | Right by 30 degrees | Factoring out 3 gives sin 3(x – 30 degrees), and a minus means 30 degrees to the right. |
| 10 | The period 120 degrees is right, but the amplitude should be 2 | Amplitude comes from the outside factor 2 while the period comes from the 3 on x, and the two cannot be swapped. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
For reverse questions, write amplitude and centre first, then the multiplier of x.Tutor Registration →
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