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Contents · 0% complete
  1. Learning objectives
  2. Core idea
  3. Key formulae
  4. Worked examples
  5. Explore the graph
  6. Common mistakes
  7. Exam tip
  8. Guided practice
  9. Practice questions
  10. Challenge
  11. Summary
  12. Quick quiz
  13. Same topic, different board
  14. Past paper questions
Learn9FM0Option · Further Pure Mathematics 1The t-formulaeThe t-substitution
The t-formulae · Subtopic 1 of 1

The t-Formulae

With t=tan⁡x2t=\tan\frac x2: sin⁡x=2t1+t2\sin x=\frac{2t}{1+t^2}, cos⁡x=1−t21+t2\cos x=\frac{1-t^2}{1+t^2}, tan⁡x=2t1−t2\tan x=\frac{2t}{1-t^2}, and dx=2 dt1+t2dx=\frac{2\,dt}{1+t^2}. These turn any rational expression in sin⁡x\sin x and cos⁡x\cos x into an algebraic one in tt, for solving equations and integrating.

Edexcel 9FM0 Option≈ 40 minRevision 1 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Derive the tt-formulae from double-angle identities.
  • Solve trigonometric equations using tt.
  • Integrate rational functions of sin⁡x\sin x and cos⁡x\cos x with t=tan⁡x2t=\tan\frac x2.
Pause and think

How could ∫dx2+cos⁡x\int\frac{dx}{2+\cos x} possibly be done?

Section 2 of 14

Core idea

Start hereOne substitution to rule them all

Integrals like ∫dx2+cos⁡x\int\frac{dx}{2+\cos x} resist every usual trick. But set t=tan⁡x2t=\tan\frac x2 and every sine, cosine and dxdx becomes a fraction in tt. Trigonometry turns into algebra.

The big idea

t=tan⁡x2t=\tan\frac x2: sin⁡x=2t1+t2\sin x=\frac{2t}{1+t^2}, cos⁡x=1−t21+t2\cos x=\frac{1-t^2}{1+t^2}, tan⁡x=2t1−t2\tan x=\frac{2t}{1-t^2}, dx=2 dt1+t2dx=\frac{2\,dt}{1+t^2}.

1. Deriving the formulae

sin⁡x\sin x and cos⁡x\cos x in tt
  1. sin⁡x=2sin⁡x2cos⁡x2=2tan⁡x2sec⁡2x2\sin x=2\sin\tfrac x2\cos\tfrac x2=\frac{2\tan\frac x2}{\sec^2\frac x2}
    Double angle; divide top and bottom by cos⁡2x2\cos^2\frac x2.
  2. =2t1+t2=\frac{2t}{1+t^2}
    sec⁡2=1+tan⁡2\sec^2=1+\tan^2.
  3. cos⁡x=cos⁡2x2−sin⁡2x2=1−t21+t2\cos x=\cos^2\tfrac x2-\sin^2\tfrac x2=\frac{1-t^2}{1+t^2}
    Same division.
  4. x=2arctan⁡t⇒dx=21+t2dtx=2\arctan t\Rightarrow dx=\frac{2}{1+t^2}dt
    Differentiate.

2. Equations

Worked example3sin⁡x+4cos⁡x=43\sin x+4\cos x=4, 0<x<2π0<x<2\pi

Solve.

Goal
xx.
Strategy
Substitute, multiply by 1+t21+t^2.
  1. 6t+4(1−t2)=4(1+t2)6t+4(1-t^2)=4(1+t^2)
    Clear denominators.
  2. 8t2−6t=0⇒t=0, 348t^2-6t=0\Rightarrow t=0,\ \tfrac34
    Factorise.
  3. t=0⇒x=0 (excluded); x=2arctan⁡34t=0\Rightarrow x=0\text{ (excluded)};\ x=2\arctan\tfrac34
    Convert back.
  4. x=π? 3(0)+4(−1)≠4x=\pi?\ 3(0)+4(-1)\neq4
    Check the value tt can't reach.
Answerx=2arctan⁡34≈1.287x=2\arctan\frac34\approx1.287.
✓ Check3sin⁡1.287+4cos⁡1.287=2.88+1.12=43\sin1.287+4\cos1.287=2.88+1.12=4 ✓.

3. Integrals

Worked example∫0π/2dx2+cos⁡x\int_0^{\pi/2}\frac{dx}{2+\cos x}

Evaluate.

Goal
Value.
Strategy
Change limits to t=0,1t=0,1.
  1. 2+cos⁡x=2+2t2+1−t21+t2=3+t21+t22+\cos x=\frac{2+2t^2+1-t^2}{1+t^2}=\frac{3+t^2}{1+t^2}
    Common denominator.
  2. ∫011+t23+t2⋅21+t2dt=∫0123+t2dt\int_0^1\frac{1+t^2}{3+t^2}\cdot\frac{2}{1+t^2}dt=\int_0^1\frac{2}{3+t^2}dt
    The 1+t21+t^2 cancels.
  3. =23[arctan⁡t3]01=23⋅π6=\frac2{\sqrt3}\left[\arctan\frac t{\sqrt3}\right]_0^1=\frac2{\sqrt3}\cdot\frac\pi6
    arctan integral.
Answerπ33≈0.605\frac\pi{3\sqrt3}\approx0.605.
✓ CheckIntegrand between 13\frac13 and 12\frac12 over width 1.571.57: between 0.520.52 and 0.790.79 ✓.
Common misconceptionLost x=πx=\pi Always test x=πx=\pi (and odd multiples) separately.

Summary

  • Derive from double angles.
  • dx=2dt1+t2dx=\frac{2dt}{1+t^2}.
  • Check x=πx=\pi.
Key words
Weierstrass substitution
t=tan⁡x2t=\tan\frac x2.
HistoryWeierstrass substitution. Named after Karl Weierstrass, though Euler used it a century earlier (1768). Spivak called it "the world's sneakiest substitution".
Fun factRational points. Rational tt gives rational (cos⁡x,sin⁡x)(\cos x,\sin x): every Pythagorean triple, like (3,4,5)(3,4,5) from t=12t=\frac12, comes from this formula.
Section 3 of 14

Key formulae

t-formulaet=tan⁡x2:sin⁡x=2t1+t2, cos⁡x=1−t21+t2, tan⁡x=2t1−t2t=\tan\frac x2:\quad\sin x=\frac{2t}{1+t^2},\ \cos x=\frac{1-t^2}{1+t^2},\ \tan x=\frac{2t}{1-t^2}
Differentialdx=21+t2 dtdx=\frac{2}{1+t^2}\,dt
TrickCheck x = π. t=tan⁡π2t=\tan\frac\pi2 is undefined, so x=πx=\pi never appears as a tt-solution. Substitute x=πx=\pi into the original equation separately.
Section 4 of 14

Worked examples

Example 1 · An equation0 / 3 steps
Solve 3sin⁡x+4cos⁡x=43\sin x+4\cos x=4, 0<x<2π0<x<2\pi.
    Example 2 · An integral0 / 3 steps
    Evaluate ∫0π/2dx1+sin⁡x\int_0^{\pi/2}\frac{dx}{1+\sin x}.
      Section 5 of 14

      Explore the graph

      t = tan(x/2) on the unit circle
      Loading the interactive graph…
      Try this
      1. The line of gradient tt from (−1,0)(-1,0) meets the circle at (cos⁡x,sin⁡x)(\cos x,\sin x).
      2. Its angle at (−1,0)(-1,0) is x2\frac x2: that's why t=tan⁡x2t=\tan\frac x2.

      Every point of the circle (except (−1,0)(-1,0)) corresponds to one tt.

      In the real worldComputer algebra. Symbolic integrators use this substitution as a fallback to guarantee they can integrate any rational function of sine and cosine.
      Section 6 of 14

      Common mistakes

      Incorrectt=tan⁡xt=\tan x
      Correctt=tan⁡x2t=\tan\frac x2
      The half-angle makes the formulae work.
      IncorrectForgetting dx=2dt1+t2dx=\frac{2dt}{1+t^2}
      CorrectInclude it
      The substitution changes the measure too.
      IncorrectMissing x=πx=\pi
      CorrectCheck separately
      tt is undefined there.

      ∫ 1/(2 + cos x) dx

      t = tan(x/2): the world's sneakiest substitution.

      Section 7 of 14

      Exam tip

      Exam tipChange the limits. Convert the limits to tt values (x=π2→t=1x=\frac\pi2\to t=1) so you never need to go back to xx.
      Section 8 of 14

      Guided practice

      SA-FP-TF-G011 markL1 FoundationNon-calculatorIdentities
      If t=tan⁡x2=1t=\tan\frac x2=1, find sin⁡x\sin x.

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      Salam Academy Original Practice · Proposed mark scheme
      Section 9 of 14

      Practice questions

      SA-FP-TF-0012 marksL1 FoundationNon-calculatorIdentities
      If t=12t=\frac12, find cos⁡x\cos x.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-TF-0021 markL1 FoundationNon-calculatorIntegrals
      With t=tan⁡x2t=\tan\frac x2, dxdt=k1+t2\frac{dx}{dt}=\frac{k}{1+t^2}. Find kk.

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-TF-0034 marksL3 ExamCalculatorEquations
      Solve 3sin⁡x+4cos⁡x=43\sin x+4\cos x=4 for 0<x<π0<x<\pi (radians, 4 s.f.).

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      Salam Academy Original Practice · Proposed mark scheme
      SA-FP-TF-0044 marksL3 ExamNon-calculatorIntegrals
      Evaluate ∫0π/2dx1+sin⁡x\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sin x}.

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      Salam Academy Original Practice · Proposed mark scheme

      More questions on every skill in this subtopic are in the question bank under Practise.

      Section 10 of 14

      Challenge

      SA-FP-TF-0055 marksL4 ChallengingCalculatorIntegrals
      Evaluate ∫0π/2dx2+cos⁡x\displaystyle\int_0^{\pi/2}\frac{dx}{2+\cos x} (to 4 s.f.).

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      Salam Academy Original Practice · Proposed mark scheme
      Section 11 of 14

      Summary

      • sin⁡\sin, cos⁡\cos, tan⁡\tan in tt.
      • dx=2dt1+t2dx=\frac{2dt}{1+t^2}.
      • Check x=πx=\pi.
      Section 12 of 14

      Quick quiz

      Q1 · SA-FP-TF-Q011 markL1 FoundationNon-calculatorIdentities
      With t=tan⁡x2t=\tan\frac x2, tan⁡x=\tan x=
      Choose one option

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      Salam Academy Original Practice · Proposed mark scheme
      Section 13 of 14

      Same topic, different board

      Specification
      A level Further Mathematics 9FM0
      Where
      Further Pure 1 · The t-formulae
      Expected
      Derive and use sin x = 2t/(1+t²), cos x = (1−t²)/(1+t²), tan x = 2t/(1−t²) with t = tan(x/2); solve equations, prove identities and integrate.
      Calculator
      Permitted; show enough working to earn the method marks.
      Typical style
      Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
      Open in
      9FM0 · Option
      Section 14 of 14

      Past paper questions

      No past paper questions are tagged to this subtopic yet. We add papers every week.

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      Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel 9FM0, First teaching 2017 (official specification).