Modelling using t-formulae (Edexcel A Level Further Maths): Revision Note

Exam code: 9FM0

Author

Mark Curtis

Last updated

5 January 2026

Modelling using t-formulae

What are the t-formulae?

The three t-formulae state that if $t = \tan \frac{θ}{2}$ then
- $\sin θ = \frac{2 t}{1 + t^{2}}$
- $\cos θ = \frac{1 - t^{2}}{1 + t^{2}}$
- $\tan θ = \frac{2 t}{1 - t^{2}}$
They express $\sin θ$ , $\cos θ$ and $\tan θ$ in terms of one variable only, $t$
From these, you can see the reciprocals
- $cosec θ = \frac{1 + t^{2}}{2 t}$
- $\sec θ = \frac{1 + t^{2}}{1 - t^{2}}$
- $\cot θ = \frac{1 - t^{2}}{2 t}$

Examiner Tips and Tricks

You must learn the t-formulae for $\sin θ$ , $\cos θ$ and $\tan θ$ as they are not given in the formulae booklet!

How do I model situations using the t-formulae?

The t-formulae can be used to investigate mathematical models
- Use $t = \tan \frac{θ}{2}$ to rewrite the model in terms of $t$
  - This gives algebraic fractions in $t$
  - which can be simplified by adding, dividing, etc
  - and often factorised
You may need to adapt the t-substitution to match the model, e.g.:
- for models in $\sin 4 θ$ and $\cos 4 θ$ use $t = \tan 2 θ$
- for models in $\tan \frac{x}{3}$ and $\sin \frac{x}{3}$ use $t = \tan \frac{x}{6}$

How do I find derivatives in terms of t?

Always differentiate the original equation first, before substituting in $t$
- e.g. the model $h = \sin 8 x + \cos 8 x$ can be written in terms of $t = \tan 4 x$
  - $h = \frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}} = \frac{1 + 2 t - t^{2}}{1 + t^{2}}$
- but to find $\frac{d h}{d x}$ from this would involve a complicated chain rule (with quotient rule)
  - $\frac{d h}{d x} = \frac{d h}{d t} \times \frac{d t}{d x} = \frac{d}{d t} (\frac{1 + 2 t - t^{2}}{1 + t^{2}}) \times \frac{d}{d x} (\tan 4 x)$
- so instead go back to the original equation $h = \sin 8 x + \cos 8 x$ and find $\frac{d h}{d x}$ of this
  - $\frac{d h}{d x} = 8 \cos 8 x - 8 \sin 8 x$
- then substitute in $t = \tan 4 x$ at the end
  - $\frac{d h}{d x} = 8 (\frac{1 - t^{2}}{1 + t^{2}}) - 8 (\frac{2 t}{1 + t^{2}}) = \frac{8 (1 - 2 t - t^{2})}{1 + t^{2}}$

Worked Example

The amplitude, $A$ metres, of part of a wave that varies with distance, $x$ metres, is modelled by

$A = 1 + 4 \sin \frac{x}{2} - \cos x 0 < x < 2 π$

(a) Show that

$\frac{d A}{d x} = \frac{2 (1 - t) {(1 + t)}^{3}}{{(1 + t^{2})}^{2}}$

where $t = \tan \frac{x}{4}$ .

Answer:

Differentiate $A = 1 + 4 \sin \frac{x}{2} - \cos x$ with respect to $x$ (avoid substituting $t = \tan \frac{x}{4}$ into $A$ first)

$\frac{d A}{d x} = 2 \cos \frac{x}{2} + \sin x$

Now convert $2 \cos \frac{x}{2} + \sin x$ into $t$ -formulae

If $t = \tan \frac{x}{4}$ then $\sin \frac{x}{2} = \frac{2 t}{1 + t^{2}}$ and $\cos \frac{x}{2} = \frac{1 - t^{2}}{1 + t^{2}}$

$\cos \frac{x}{2} = \frac{1 - t^{2}}{1 + t^{2}}$

To find $\sin x$ in terms of $t$ , use the double-angle formula $\sin 2 A \equiv 2 \sin A \cos A$

$\begin{array}{rcl} \sin x & \equiv & 2 \sin \frac{x}{2} \cos \frac{x}{2} \\ = & 2 (\frac{2 t}{1 + t^{2}}) (\frac{1 - t^{2}}{1 + t^{2}}) \\ = & \frac{4 t (1 - t^{2})}{{(1 + t^{2})}^{2}} \end{array}$

Substitute $\cos \frac{x}{2}$ and $\begin{array}{rcl} \sin x \end{array}$ into $\frac{d A}{d x}$

$\frac{d A}{d x} = 2 (\frac{1 - t^{2}}{1 + t^{2}}) + \begin{array}{rcl} \frac{4 t (1 - t^{2})}{{(1 + t^{2})}^{2}} \end{array}$

Add the algebraic fractions using a common denominator of ${(1 + t^{2})}^{2}$

$\frac{d A}{d x} = \frac{2 (1 - t^{2}) (1 + t^{2}) + 4 t (1 - t^{2})}{{(1 + t^{2})}^{2}}$

Factorise out $2 (1 - t^{2})$ from the numerator

$\frac{d A}{d x} = \frac{2 (1 - t^{2}) [(1 + t^{2}) + 2 t]}{{(1 + t^{2})}^{2}}$

Rearrange $(1 + t^{2}) + 2 t$ to $1 + 2 t + t^{2}$ , which factorises to ${(1 + t)}^{2}$

$\frac{d A}{d x} = \frac{2 (1 - t^{2}) {(1 + t)}^{2}}{{(1 + t^{2})}^{2}}$

To make this look like the answer in the question, use the difference of two squares to write $1 - t^{2}$ as $(1 + t) (1 - t)$

$\frac{d A}{d x} = \frac{2 (1 + t) (1 - t) {(1 + t)}^{2}}{{(1 + t^{2})}^{2}}$

Now combine the $(1 + t)$ and ${(1 + t)}^{2}$ in the numerator into one single power

$\frac{d A}{d x} = \frac{2 (1 - t) {(1 + t)}^{3}}{{(1 + t^{2})}^{2}}$

(b) Given that

$\frac{d^{2} A}{d x^{2}} = \frac{{(1 + t)}^{2} (t^{2} - 4 t + 1)}{{(1 + t^{2})}^{2}}$

determine whether the model represents a peak (crest) or a trough (dip) of the wave.

Answer:

A peak or trough means a maximum or a minimum point, which both occur when $\frac{d A}{d x} = 0$

Find the value of $t$ that makes $\frac{d A}{d x} = 0$ using the answer in part (a)

$\frac{2 (1 - t) {(1 + t)}^{3}}{{(1 + t^{2})}^{2}} = 0$

Setting the numerator equal to zero gives

$t = 1 or t = - 1$

Check to see if these solutions are within the range $0 < x < 2 π$ given in the question

For $t = \tan \frac{x}{4} = 1$ there is one possible solution in the range

$\begin{array}{rcl} \frac{x}{4} & = & artan (1) + n π \\ \frac{x}{4} & = & \frac{π}{4} + n π \\ x & = & π + 4 n π \\ x & = & . . . - 3 π, π, 5 π, . . . \end{array}$

$\begin{array}{rcl} x & = & π \end{array}$

For $t = \tan \frac{x}{4} = - 1$ there are no possible solutions in the range

$\begin{array}{rcl} \frac{x}{4} & = & artan (- 1) + n π \\ \frac{x}{4} & = & - \frac{π}{4} + n π \\ x & = & - π + 4 n π \\ x & = & . . ., - π, 3 π, . . . \end{array}$

This means the maximum or minimum is at $x = π$ when $t = 1$

To find out the nature of the stationary point, substitute $t = 1$ into $\frac{d^{2} A}{d x^{2}}$

$\begin{array}{rcl} \frac{d^{2} A}{d x^{2}} & = & \frac{{(1 + 1)}^{2} (1^{2} - 4 \times 1 + 1)}{{(1 + 1^{2})}^{2}} \\ = & \frac{2^{2} \times (- 2)}{2^{2}} < 0 \end{array}$

The second derivative is negative, so $x = π$ is a maximum point

The model represents a peak (crest) of the wave

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Modelling using t-formulae (Edexcel A Level Further Maths): Revision Note

Modelling using t-formulae

What are the t-formulae?

How do I model situations using the t-formulae?

How do I find derivatives in terms of t?

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Further Trigonometry

The t-formulae

Trig identities using t-formulae

Trig Equations using t-formulae

Modelling using t-formulae

Further Calculus

Taylor Series

Taylor Series

Limits using Series

Series Solutions to Differential Equations

Methods in Calculus

Leibnitz's Theorem

L'Hospital's Rule

The Weierstrass Substitution

Further Differential Equations

Reducing Differential Equations

Reducing First-Order Differential Equations

Reducing Second-Order Differential Equations

Coordinate Systems

Properties of Ellipses, Parabolas & Hyperbolas

Properties of Ellipses

Properties of Parabolas

Properties of Hyperbolas

Properties of Rectangular Hyperbolas

Tangent & Normals to Ellipses, Parabolas & Hyperbolas

Tangents & Normals to Ellipses

Tangents & Normals to Parabolas

Tangents & Normals to Hyperbolas

Tangents & Normals to Rectangular Hyperbolas

Loci Problems

Loci Problems

Further Vectors

Further Vectors

Vector (Cross) Product

Scalar Triple Product

Vector Geometry with Points, Lines & Planes

Shortest Distances using Vector Rules

Areas & Volumes using Vector Rules

Direction Ratios & Direction Cosines

Further Numerical Methods

Further Numerical Methods

Numerical Solutions of First-Order Differential Equations

Numerical Solutions of Second-Order Differential Equations

Simpson's Rule

Inequalities

Inequalities

Solving Inequalities involving Algebraic Fractions

Solving Inequalities involving Modulus Functions

Author: Mark Curtis