Use double-angle identities without losing valid solutions
Choose an identity that suits the available ratio, expand the solution interval and protect values lost by division or a narrower substitution.
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
- Use compound-angle formulae, factor products and solve basic trigonometric equations.
- Check original domains and distinguish sin² θ from sin 2θ.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which identity is correct for sin 2θ?
What you will learn
- Use sin 2A, all three forms of cos 2A and tan 2A with their valid domains.
- Solve doubled-angle equations over the complete transformed interval without deleting a valid zero factor.
Choose a form that uses the ratio you know
Putting the same angle into the addition formulae gives sin 2A = 2 sin A cos A and cos 2A = cos² A − sin² A. Using sin² A + cos² A = 1 produces cos 2A = 1 − 2 sin² A = 2 cos² A − 1. If sin A is supplied, choose the sine-square form; if cosine is supplied, choose the cosine-square form. For sin A = 3/5, cos 2A = 1 − 18/25 = 7/25 without needing a sign choice for cos A.
The tangent form is tan 2A = 2 tan A/(1 − tan² A) where both tan A and tan 2A are defined. Its denominator excludes tan A = ±1. It is a representation on a restricted domain, not a reason to exclude extra inputs from an original tan 2A equation: at A = 90°, tan A is undefined, but tan 180° = 0 is valid. If the substituted representation narrows the original domain, inspect those omitted inputs separately.
Factor before dividing and transform the whole interval
For sin 2θ = cos θ, use 2 sin θ cos θ = cos θ and factor cos θ(2 sin θ − 1) = 0. On 0° ≤ θ < 360°, the cosine factor gives 90° and 270°, while sin θ = 1/2 gives 30° and 150°. Dividing by cos θ at the start would erase two valid solutions. Checking the original equation distinguishes a valid zero factor from an excluded original denominator.
If cos 2θ = 1/2 on 0° ≤ θ < 360°, the doubled angle runs through 0° ≤ 2θ < 720°. Its four values are 60°, 300°, 420° and 660°, giving θ = 30°, 150°, 210° and 330°. Searching only one turn of 2θ would miss two answers. State whether an interval includes its endpoints, then filter the recovered angles using the original interval and equation.
Pause and explain
Why must cos θ = 0 be checked before dividing sin 2θ = cos θ by cos θ?
Worked example
Solve sin 2θ = cos θ for 0° ≤ θ < 360°.
Show the worked solution
- Use sin 2θ = 2 sin θ cos θ and bring both terms to one side.
- Factor cos θ(2 sin θ − 1) = 0 without dividing by the cosine factor.
- Use both factors: cos θ = 0 gives 90° and 270°, while sin θ = 1/2 gives 30° and 150°; check all four in the original equation.
Answer θ = 30°, 90°, 150° or 270°.
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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.
Solve cos 2θ = 1/2 for 0° ≤ θ < 360°.
Give me a hint
Double both limits before listing the cosine solutions, then divide every angle by 2.
Compare my reasoning
- The new angle u = 2θ satisfies 0° ≤ u < 720°.
- Cosine 1/2 occurs at u = 60°, 300°, 420° and 660° in that interval.
- Divide by 2 to obtain θ = 30°, 150°, 210° and 330°, all satisfying the original equation.
θ = 30°, 150°, 210° or 330°.
Look for these in your work
- Search the whole doubled interval rather than only one turn.
- Return every solution to the original variable and interval.
Solve tan 2θ = 0 for 0° ≤ θ < 360°. Explain why using tan θ as a variable alone can miss answers.
Give me a hint
Solve the original doubled-angle equation first, then inspect the domain of the proposed auxiliary tangent.
Compare my reasoning
- On 0° ≤ 2θ < 720°, tangent is zero at 0°, 180°, 360° and 540°.
- Hence θ = 0°, 90°, 180° and 270°; tan 2θ is defined and zero at each.
- The auxiliary tan θ is undefined at 90° and 270°, so its quotient formula alone would miss those valid original inputs. θ = 360° is excluded by the stated upper endpoint.
θ = 0°, 90°, 180° or 270°; 90° and 270° require checking outside the tan θ substitution's domain.
Look for these in your work
- Check the original doubled-angle function's domain.
- Recover valid inputs omitted by a narrower auxiliary expression.
A supplied angular response is B(θ) = 2 sin 2θ − cos θ for 0 ≤ θ < 2π radians. Find every zero exactly, using α = sin⁻¹(1/4).
Give me a hint
Use the double-angle sine formula and retain both factors instead of dividing by cosine.
Compare my reasoning
- Rewrite B(θ) as 4 sin θ cos θ − cos θ = cos θ(4 sin θ − 1).
- The cosine zeros are π/2 and 3π/2; the other factor gives sin θ = 1/4 and angles α and π − α.
- All four lie in the interval and satisfy the original response; α is the positive principal inverse-sine angle.
θ = α, π/2, π − α or 3π/2, where α = sin⁻¹(1/4).
Look for these in your work
- Select the useful identity and preserve every zero factor.
- Use the radian interval and verify all recovered roots.
Common mistakes
- sin 2θ is not 2 sin θ or sin² θ.
- A zero factor may be a valid solution; division by that factor can delete it.
- A tangent substitution can omit angles where tan θ is undefined but tan 2θ is defined; check them separately.
What can you explain now?
If sin θ = 3/5, find cos 2θ exactly.
Compare with the explanation
7/25.
Use 1 − 2 sin² θ = 1 − 18/25, so no cosine sign choice is required.
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Make it stick
Give the double-angle identities and choose a cosine form for a supplied sine value. Explain two ways an algebraic step can accidentally delete original equation solutions.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.