Compound Angle Formulae (DP IB Analysis & Approaches (AA)): Revision Note

Amber

Written by: Amber

Reviewed by: Dan Finlay

Updated on

Compound angle formulae

What are the compound angle formulae?

  • The compound angle formulae are:

    • sin left parenthesis A space plus-or-minus space B right parenthesis space equals space sin A cos B space plus-or-minus space cos A sin B

      • The sign is the same on both sides

    • cos left parenthesis A space plus-or-minus space B right parenthesis space equals space cos A cos B space minus-or-plus space sin A sin B

      • The sign changes

    • tan open parentheses A space plus-or-minus space B close parentheses blank equals space fraction numerator tan A space plus-or-minus space tan B over denominator 1 space minus-or-plus space tan A tan B end fraction

      • The sign is the same on the numerator

      • But changes on the denominator

Examiner Tips and Tricks

You are given these formulas in the formula booklet. Be careful with the signs.

How can I use the compound angle formulas to find exact trig values?

  • STEP 1
    Write the angle as a sum or difference of multiples of 30° or 45°

    • e.g. 15° = 45° - 30°

  • STEP 2
    Use the relevant compound angle formula

    • e.g. sin open parentheses 45 minus 30 close parentheses equals sin 45 cos 30 minus cos 45 sin 30

  • STEP 3
    Evaluate the known trig values and simplify

    • e.g. sin 15 equals fraction numerator square root of 2 over denominator 2 end fraction cross times fraction numerator square root of 3 over denominator 2 end fraction minus fraction numerator square root of 2 over denominator 2 end fraction cross times 1 half equals fraction numerator square root of 6 minus square root of 2 over denominator 2 end fraction

How can I use the compound angle formulas?

  • You can use the compound angle formulas to prove identities are true

  • You can use the formulas to help you solve equations

Worked Example

a) Show that begin mathsize 16px style tan invisible function application open parentheses x plus pi over 4 close parentheses minus blank tan invisible function application open parentheses x minus pi over 4 close parentheses equals blank fraction numerator 2 left parenthesis tan squared invisible function application x plus 1 right parenthesis over denominator 1 minus tan squared invisible function application x end fraction blank end style

3-6-2-ib-aa-hl-c-a-form-we-solution-a

b) Hence, solve begin mathsize 16px style space tan invisible function application open parentheses x plus pi over 4 close parentheses minus blank tan invisible function application open parentheses x minus pi over 4 close parentheses equals blank minus 4 end style  for begin mathsize 16px style 0 blank less or equal than x blank less or equal than blank pi over 2 end style

3-6-2-ib-aa-hl-c-a-form-we-solution-b

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Amber

Author: Amber

Expertise: Maths Content Creator

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject 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.