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Trigonometric Addition Formulae (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

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 open parentheses A plus B close parentheses equals sin A cos B plus cos A sin B

    • sin open parentheses A minus B close parentheses equals sin A cos B minus 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 open parentheses A plus B close parentheses equals cos A cos B minus sin A sin B

    • cos open parentheses A minus B close parentheses equals cos A cos B plus 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 invisible function application open parentheses A plus B close parentheses blank identical to fraction numerator tan invisible function application A plus tan invisible function application B over denominator 1 minus tan invisible function application A tan invisible function application B end fraction

    • tan invisible function application open parentheses A minus B close parentheses blank identical to fraction numerator tan invisible function application A minus tan invisible function application B over denominator 1 plus tan invisible function application A tan invisible function application B end fraction

      • 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 bold italic A bold equals bold italic B in the '+' versions of the addition formulae 

  • The sin version is

    • sin open parentheses 2 A close parentheses equals 2 sin A cos A

  • The cos version is

    • cos open parentheses 2 A close parentheses equals cos squared A minus sin squared A equals 1 minus 2 sin squared A equals 2 cos squared A minus 1

      • The last two forms come from using the sin squared A plus cos squared A equals 1 identity

      • i.e. cos squared A equals 1 minus sin squared A and sin squared A equals 1 minus cos squared A

  • The tan version is

    • tan open parentheses 2 A close parentheses equals fraction numerator 2 tan A over denominator 1 minus tan squared A end fraction

  • 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 open parentheses x plus pi over 4 close parentheses minus blank tan open parentheses x minus pi over 4 close parentheses equals fraction numerator 2 left parenthesis tan squared x plus 1 right parenthesis over denominator 1 minus tan squared x end fraction.

Use the trigonometric addition formulae for tan

Also recall that tan pi over 4 equals 1

tan open parentheses x plus pi over 4 close parentheses equals fraction numerator tan x plus tan pi over 4 over denominator 1 minus tan x tan pi over 4 end fraction equals fraction numerator tan x plus 1 over denominator 1 minus tan x end fraction

tan open parentheses x minus pi over 4 close parentheses equals fraction numerator tan x minus tan pi over 4 over denominator 1 plus tan x tan pi over 4 end fraction equals fraction numerator tan x minus 1 over denominator 1 plus tan x end fraction

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

table row cell tan open parentheses x plus pi over 4 close parentheses minus tan open parentheses x minus pi over 4 close parentheses end cell equals cell fraction numerator tan x plus 1 over denominator 1 minus tan x end fraction minus fraction numerator tan x minus 1 over denominator 1 plus tan x end fraction end cell row blank equals cell fraction numerator open parentheses tan x plus 1 close parentheses open parentheses 1 plus tan x close parentheses minus open parentheses tan x minus 1 close parentheses open parentheses 1 minus tan x close parentheses over denominator open parentheses 1 minus tan x close parentheses open parentheses 1 plus tan x close parentheses end fraction end cell row blank equals cell fraction numerator tan squared x plus 2 tan x plus 1 plus tan squared x minus 2 tan x plus 1 over denominator 1 minus tan squared x end fraction end cell row blank equals cell fraction numerator 2 tan squared x plus 2 over denominator 1 minus tan squared x end fraction end cell row blank equals cell fraction numerator 2 open parentheses tan squared x plus 1 close parentheses over denominator 1 minus tan squared x end fraction end cell end table

bold tan stretchy left parenthesis x plus pi over 4 stretchy right parenthesis bold minus blank bold tan stretchy left parenthesis x minus pi over 4 stretchy right parenthesis bold equals fraction numerator bold 2 bold left parenthesis bold tan to the power of bold 2 bold invisible function application bold x bold plus bold 1 bold right parenthesis over denominator bold 1 bold minus bold tan to the power of bold 2 bold invisible function application bold x end fraction

b) Hence solve begin mathsize 16px style space tan open parentheses x plus pi over 4 close parentheses minus tan open parentheses x minus pi over 4 close parentheses equals negative 4 end style  in the interval  0 blank less or equal than x blank less or equal than blank pi over 2.

Substitute the result from part (a) into the equation

Then rearrange and solve

table row cell fraction numerator 2 left parenthesis tan squared x plus 1 right parenthesis over denominator 1 minus tan squared x end fraction end cell equals cell negative 4 end cell row cell fraction numerator tan squared x plus 1 over denominator 1 minus tan squared x end fraction end cell equals cell negative 2 end cell row cell tan squared x plus 1 end cell equals cell 2 tan squared x minus 2 end cell row cell tan squared x end cell equals 3 row cell tan x end cell equals cell plus-or-minus square root of 3 end cell row x equals cell pi over 3 comma space minus pi over 3 end cell end table


But only pi over 3 is in the range 0 less or equal than x less or equal than pi over 2

bold italic x bold equals bold pi over bold 3

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.