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

Exam code: 9FM0

Author

Mark Curtis

Last updated

5 January 2026

Trig identities 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 derive the t-formulae?

To derive $\sin θ = \frac{2 t}{1 + t^{2}}$
- first use the double-angle formula $\sin 2 A \equiv 2 \sin A \cos A$
  - $\sin θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2}$
- then multiply by $\frac{\sec^{2} \frac{θ}{2}}{\sec^{2} \frac{θ}{2}}$ and use $1 + \tan^{2} A \equiv \sec^{2} A$ in the denominator
  - $\sin θ \equiv 2 \sin \frac{θ}{2} \cos \frac{θ}{2} \times \frac{\sec^{2} \frac{θ}{2}}{\sec^{2} \frac{θ}{2}}$
  - $\sin θ \equiv \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2} \sec^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}$
- Simplify the numerator, $2 s c \times \frac{1}{c^{2}} = \frac{2 s}{c} = 2 t$
  - $\sin θ \equiv \frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}} = \frac{2 t}{1 + t^{2}}$
To derive $\cos θ = \frac{1 - t^{2}}{1 + t^{2}}$
- first use the double-angle formula $\cos 2 A \equiv \cos^{2} A - \sin^{2} A$
  - $\cos θ \equiv \cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}$
- then multiply by $\frac{\sec^{2} \frac{θ}{2}}{\sec^{2} \frac{θ}{2}}$ and use $1 + \tan^{2} A \equiv \sec^{2} A$ in the denominator
  - $\cos θ \equiv (\cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}) \times \frac{\sec^{2} \frac{θ}{2}}{\sec^{2} \frac{θ}{2}}$
  - $\cos θ \equiv \frac{(\cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}) \sec^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}$
- Simplify the numerator, $(c^{2} - s^{2}) \times \frac{1}{c^{2}} = 1 - \frac{s^{2}}{c^{2}} = 1 - t^{2}$
  - $\cos θ \equiv \frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}} = \frac{1 - t^{2}}{1 + t^{2}}$
To derive $\tan θ = \frac{2 t}{1 - t^{2}}$
- either use the double-angle formula $\tan 2 A \equiv \frac{2 \tan A}{1 - \tan^{2} A}$
  - $\tan θ \equiv \frac{2 \tan \frac{θ}{2}}{1 - \tan^{2} \frac{θ}{2}} = \frac{2 t}{1 - t^{2}}$
- or substitute the results for $\sin θ$ and $\cos θ$ above into $\tan θ \equiv \frac{\sin θ}{\cos θ}$
  - $\tan θ \equiv \frac{\sin θ}{\cos θ} = \frac{2 t}{1 + t^{2}} \div \frac{1 - t^{2}}{1 + t^{2}} = \frac{2 t}{1 + t^{2}} \times \frac{1 + t^{2}}{1 - t^{2}} = \frac{2 t}{1 - t^{2}}$

How do I prove trigonometric identities using the t-formulae?

To prove trigonometric identities using t-formulae
- let $t = \tan \frac{θ}{2}$
- convert the trig functions on one or both sides of the identity into algebraic fractions in $t$ using
  - $\sin θ = \frac{2 t}{1 + t^{2}}$ , $\cos θ = \frac{1 - t^{2}}{1 + t^{2}}$ , $\tan θ = \frac{2 t}{1 - t^{2}}$
  - $cosec θ = \frac{1 + t^{2}}{2 t}$ , $\sec θ = \frac{1 + t^{2}}{1 - t^{2}}$ , $\cot θ = \frac{1 - t^{2}}{2 t}$
- rearrange the algebraic fractions in $t$ to prove the identity
  - e.g. adding, subtracting, multiplying, dividing
- convert back to $θ$ for the final step
  - using $t = \tan \frac{θ}{2}$

Examiner Tips and Tricks

You may need to adapt the t-substitution to match the identity, e.g.:

for identities in $\sin 4 θ$ and $\cos 4 θ$ use $t = \tan 2 θ$
for identities in $\tan \frac{x}{3}$ and $\sin \frac{x}{3}$ use $t = \tan \frac{x}{6}$

Worked Example

Use the substitution $t = \tan \frac{θ}{2}$ to prove the identity

$\sec θ + \tan θ \equiv \frac{1 + \tan \frac{θ}{2}}{1 - \tan \frac{θ}{2}}$

for $θ \neq \frac{π}{2} + n π$ and $θ \neq π + 2 n π$ , where $n \in ℤ$ .

Answer:

Write down $\cos θ$ in terms of its $t$ -formula

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

Find $\sec θ$ by finding the reciprocal of both sides

$\sec θ = \frac{1 + t^{2}}{1 - t^{2}}$

Write down $\tan θ$ in terms of its $t$ -formula

$\tan θ = \frac{2 t}{1 - t^{2}}$

Substitute $\sec θ = \frac{1 + t^{2}}{1 - t^{2}}$ and $\tan θ = \frac{2 t}{1 - t^{2}}$ into the left-hand side of the identity

$L H S = \frac{1 + t^{2}}{1 - t^{2}} + \frac{2 t}{1 - t^{2}}$

Add the algebraic fractions

$L H S = \frac{1 + t^{2} + 2 t}{1 - t^{2}}$

The numerator rearranges to $1 + 2 t + t^{2}$ , which factorises to ${(1 + t)}^{2}$

The denominator is the difference of two squares, $(1 + t) (1 - t)$

Cancel the common factors

$\begin{array}{rcl} L H S & = & \frac{{(1 + t)}^{2}}{(1 + t) (1 - t)} \\ = & \frac{(1 + t) (1 + t)}{(1 + t) (1 - t)} \\ = & \frac{1 + t}{1 - t} \end{array}$

Convert from $t$ back to $θ$ using $t = \tan \frac{θ}{2}$

$\begin{array}{rcl} L H S & = & \frac{1 + \tan \frac{θ}{2}}{1 - \tan \frac{θ}{2}} \end{array}$

This expression is the correct right-hand side

$\begin{array}{rcl} L H S & = & \frac{1 + \tan \frac{θ}{2}}{1 - \tan \frac{θ}{2}} = R H S \end{array}$

This means

$\sec θ + \tan θ \equiv \frac{1 + \tan \frac{θ}{2}}{1 - \tan \frac{θ}{2}}$

Examiner Tips and Tricks

The reasons for the extra restrictions in the question are that

$θ \neq \frac{π}{2} + n π$ where $n \in ℤ$
- stops $\sec θ$ and $\tan θ$ from being undefined
- and stops the denominator $1 - \tan \frac{θ}{2}$ from being zero
$θ \neq π + 2 n π$ where $n \in ℤ$
- stops the $\tan \frac{θ}{2}$ terms being undefined

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

Trig identities using t-formulae

What are the t-formulae?

How do I derive the t-formulae?

How do I prove trigonometric identities using the t-formulae?

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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

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Further Vectors

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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