Express a sine cosine combination in amplitude phase form
Match coefficients to recover a positive amplitude and a correctly signed phase in either sine or cosine form.
Section 2.3 · Trigonometry
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.3 and 3.3 cover all six trigonometric functions, reciprocal squared identities, compound and double angles, and amplitude-phase forms used in simplification, exact evaluation and equations. State angle units, keep original domains and find every solution in the supplied interval. Written practice is self-checked.
Before you begin
- Expand compound-angle sine and cosine and use quadrant signs.
- Find a vector magnitude and keep degree and radian phase values consistent.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the amplitude of 3 sin θ + 4 cos θ?
What you will learn
- Express a sin θ + b cos θ as R sin(θ ± α) or R cos(θ ± β) by matching coefficients.
- Use the amplitude and phase to find a range and extrema, including negative coefficients.
Coefficient matching determines the amplitude and the phase
Expand R sin(θ + α) as R cos α sin θ + R sin α cos θ. To match a sin θ + b cos θ, require R cos α = a and R sin α = b. Squaring and adding gives R = √(a² + b²), chosen positive when the coefficients are not both zero. The phase must satisfy both cos α = a/R and sin α = b/R. A tangent inverse alone can select the wrong quadrant when a or b is negative, so check both signs. If a = b = 0 the expression is the constant zero and a unique phase is not determined.
For 3 sin θ + 4 cos θ, R = 5 and an acute phase α has cos α = 3/5 and sin α = 4/5. Hence the sine form is 5 sin(θ + α), with α = arctan(4/3) ≈ 53.1° in degrees. A cosine form is 5 cos(θ − β), where cos β = 4/5, sin β = 3/5 and β = arctan(3/4) ≈ 36.9°. Expanding either form recovers the original coefficients. Use the unrounded phase for any later calculation.
A phase changes where an extremum occurs, not its height
For −3 sin θ + 4 cos θ, choose 5 cos(θ + β) with cos β = 4/5 and sin β = 3/5: cosine addition produces 4 cos θ − 3 sin θ. Likewise 5 sin θ − 12 cos θ = 13 sin(θ − α) with cos α = 5/13 and sin α = 12/13. A minus or plus phase sign is determined by expansion; it is not chosen just because an original coefficient looks negative. These sine and cosine representations have the same all-real angle domain as the original sum.
The range of 3 sin θ + 4 cos θ is [−5, 5] over a full turn. In its cosine form, the maximum occurs at θ = β and the minimum at θ = β + 180° for 0° ≤ θ < 360°. Adding a constant 7 changes the range to [2, 12] but not the extrema's angle locations or the 360° period. If the supplied interval covers only part of a turn, check whether those extrema lie inside it rather than assume the full-period range is attained.
Pause and explain
Which form recovers −3 sin θ + 4 cos θ when cos β = 4/5 and sin β = 3/5?
Worked example
Express 3 sin θ + 4 cos θ in both sine and cosine amplitude-phase forms, with degree input θ.
Show the worked solution
- The amplitude is R = √(3² + 4²) = 5.
- For the sine form, match cos α = 3/5 and sin α = 4/5, giving 5 sin(θ + α) with α = arctan(4/3).
- For the cosine form, match cos β = 4/5 and sin β = 3/5, giving 5 cos(θ − β) with β = arctan(3/4); expand to check both coefficients.
Answer 5 sin(θ + α), α = arctan(4/3); also 5 cos(θ − β), β = arctan(3/4). In degrees α ≈ 53.1° and β ≈ 36.9°.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Express 5 sin θ − 12 cos θ as R sin(θ − α), with R > 0 and acute α, and state its full-turn range.
Give me a hint
Expand the requested sine subtraction form and match its two coefficients.
Compare my reasoning
- R cos α = 5 and R sin α = 12, so R = √(25 + 144) = 13.
- Choose cos α = 5/13 and sin α = 12/13, giving α = arctan(12/5).
- The form is 13 sin(θ − α), with full-turn range [−13, 13].
13 sin(θ − α), α = arctan(12/5); range [−13, 13].
Look for these in your work
- Match the signs in the requested expansion.
- Use the amplitude to state the range over a full cycle.
Express −3 sin θ + 4 cos θ as R cos(θ + β), with positive R and acute β. Verify by expansion.
Give me a hint
Cosine addition supplies a negative sine term automatically when β is acute.
Compare my reasoning
- Match R cos β = 4 and R sin β = 3, giving R = 5.
- Choose β = arctan(3/4), with cosine 4/5 and sine 3/5.
- Expand 5 cos(θ + β) = 4 cos θ − 3 sin θ to verify both coefficients.
5 cos(θ + β), β = arctan(3/4), with cos β = 4/5 and sin β = 3/5.
Look for these in your work
- Choose a phase sign that matches the original negative coefficient.
- Verify both terms by expanding the final expression.
A supplied response is y(θ) = 7 + 3 sin θ + 4 cos θ for 0° ≤ θ < 360°. Find its range, maximum and minimum angle locations, and period.
Give me a hint
Use the cosine form 7 + 5 cos(θ − β), with β = arctan(3/4), and keep β exact until rounding.
Compare my reasoning
- The amplitude 5 about midline 7 gives range [2, 12].
- The maximum 12 occurs at θ = β ≈ 36.9°, and the minimum 2 at θ = β + 180° ≈ 216.9°, both in the interval.
- The input is unscaled, so the period remains 360°; substitution into the exact form checks the extrema.
Range [2, 12]; maximum at θ ≈ 36.9°; minimum at θ ≈ 216.9°; period 360°.
Look for these in your work
- Separate the midline from amplitude and phase.
- Check that both extremum locations belong to the supplied interval.
Common mistakes
- The amplitude is √(a² + b²), not a + b.
- Check both phase sine and cosine signs rather than rely only on a principal tangent inverse.
- Keep the exact phase until the final rounded answer and use the same angle unit as the input.
What can you explain now?
Express 3 cos θ + 2 sin θ as R cos(θ − β), choosing acute β.
Compare with the explanation
√13 cos(θ − β), with cos β = 3/√13, sin β = 2/√13 and β = arctan(2/3).
Cosine subtraction has a positive sine product; coefficient matching gives both the phase and √13 amplitude.
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Make it stick
Expand both amplitude-phase forms to match coefficients. Explain how to choose a phase quadrant and why a vertical shift changes the range but not the extremum angle locations.
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