Trigonometric Addition Formulae (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Trigonometric Addition Formulae

What are the trigonometric addition formulae?

There are six trigonometric addition formulae (also known as compound angle formulae),
- two each for sin, cos and tan
The formulae for sin are
- $\sin (A + B) = \sin A \cos B + \cos A \sin B$
- $\sin (A - B) = \sin A \cos B - \cos A \sin B$
  - Note that the +/- sign on the left-hand side matches the one on the right-hand side

The formulae for cos are
- $\cos (A + B) = \cos A \cos B - \sin A \sin B$
- $\cos (A - B) = \cos A \cos B + \sin A \sin B$
  - Note that the +/- sign on the left-hand side is opposite to the one on the right-hand side

The formulae for tan are
- $\tan (A + B) \equiv \frac{\tan A + \tan B}{1 - \tan A \tan B}$
- $\tan (A - B) \equiv \frac{\tan A - \tan B}{1 + \tan A \tan B}$
  - Note that the +/- sign on the left-hand side matches the one in the numerator on the right-hand side, and is opposite to the one in the denominator
These formulae are all on the exam formula sheet
- so you don't need to remember them
- but you do need to be able to use them

What are the double angle formulae?

The double angle formulae are special cases of the trigonometric addition formulae
- They are formed by setting $A = B$ in the '+' versions of the addition formulae
The sin version is
- $\sin (2 A) = 2 \sin A \cos A$
The cos version is
- $\cos (2 A) = \cos^{2} A - \sin^{2} A = 1 - 2 \sin^{2} A = 2 \cos^{2} A - 1$
  - The last two forms come from using the $\sin^{2} A + \cos^{2} A = 1$ identity
  - i.e. $\cos^{2} A = 1 - \sin^{2} A$ and $\sin^{2} A = 1 - \cos^{2} A$
The tan version is
- $\tan (2 A) = \frac{2 \tan A}{1 - \tan^{2} A}$
These formulae are not on the exam formula sheet
- They are used frequently, so you may want to remember them
- But they are also easy to derive from the addition formulae that are on the sheet

How are the trigonometric addition formulae used?

The formulae can be used to find the values of trigonometric ratios without a calculator
- For example, to find the value of sin15°
  - rewrite it as sin(45–30)°
  - apply the formula for sin(A –B)
  - use your knowledge of exact values to calculate the answer
The formulae can also be used
- to derive further trigonometric identities (like the double angle formulae)
- in trigonometric proof
- to simplify trigonometric equations before solving

Examiner Tips and Tricks

Remember that the trigonometric addition formulae are on the exam formula sheet
- But always be careful with the +/- signs when using the formulae

Worked Example

a) Show that $\tan (x + \frac{π}{4}) - \tan (x - \frac{π}{4}) = \frac{2 (\tan^{2} x + 1)}{1 - \tan^{2} x}$ .

Use the trigonometric addition formulae for tan

Also recall that $\tan \frac{π}{4} = 1$

$\tan (x + \frac{π}{4}) = \frac{\tan x + \tan \frac{π}{4}}{1 - \tan x \tan \frac{π}{4}} = \frac{\tan x + 1}{1 - \tan x}$

$\tan (x - \frac{π}{4}) = \frac{\tan x - \tan \frac{π}{4}}{1 + \tan x \tan \frac{π}{4}} = \frac{\tan x - 1}{1 + \tan x}$

Substitute those into the left-hand side of the equation and rearrange

$\begin{array}{rcl} \tan (x + \frac{π}{4}) - \tan (x - \frac{π}{4}) & = & \frac{\tan x + 1}{1 - \tan x} - \frac{\tan x - 1}{1 + \tan x} \\ = & \frac{(\tan x + 1) (1 + \tan x) - (\tan x - 1) (1 - \tan x)}{(1 - \tan x) (1 + \tan x)} \\ = & \frac{\tan^{2} x + 2 \tan x + 1 + \tan^{2} x - 2 \tan x + 1}{1 - \tan^{2} x} \\ = & \frac{2 \tan^{2} x + 2}{1 - \tan^{2} x} \\ = & \frac{2 (\tan^{2} x + 1)}{1 - \tan^{2} x} \end{array}$

$\tan (x + \frac{π}{4}) - \tan (x - \frac{π}{4}) = \frac{2 (\tan^{2} x + 1)}{1 - \tan^{2} x}$

b) Hence solve $\tan (x + \frac{π}{4}) - \tan (x - \frac{π}{4}) = - 4$ in the interval $0 \leq x \leq \frac{π}{2}$ .

Substitute the result from part (a) into the equation

Then rearrange and solve

$\begin{array}{rcl} \frac{2 (\tan^{2} x + 1)}{1 - \tan^{2} x} & = & - 4 \\ \frac{\tan^{2} x + 1}{1 - \tan^{2} x} & = & - 2 \\ \tan^{2} x + 1 & = & 2 \tan^{2} x - 2 \\ \tan^{2} x & = & 3 \\ \tan x & = & \pm \sqrt{3} \\ x & = & \frac{π}{3}, - \frac{π}{3} \end{array}$

But only $\frac{π}{3}$ is in the range $0 \leq x \leq \frac{π}{2}$

$x = \frac{π}{3}$

Last updated: 20 October 2024

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