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
- $\sin (A + B) = \sin A \cos B + \cos A \sin B$
- - Note that the +/- sign on the left-hand side matches the one on the right-hand side

The formulae for cos are
- $\cos (A + B) = \cos A \cos B - \sin A \sin B$
- - 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

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
- - 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

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

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

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}

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}

Trigonometric Addition Formulae (Edexcel IGCSE Further Maths)