Applications of de Moivre's Theorem (Edexcel A Level Further Maths: Core Pure)

Revision Note

Test Yourself
Dan

Author

Dan

Expertise

Maths

Multiple Angle Formulae

de Moivre’s theorem can be applied to prove identities such as sin 5 x identical to 16 sin to the power of 5 x minus 20 sin cubed x plus 5 sin x. This allows expressions involving multiple angles to be written as a polynomial function of a single trig function which makes it easier to solve equations involving different angles.

How do I write sinkθ or coskθ in terms of powers of sinθ or cosθ?

  • STEP 1
    Use de Moivre’s theorem to write  open parentheses cos theta plus isin theta close parentheses to the power of k equals cos k theta plus isin k theta
  • STEP 2
    Use the binomial expansion to expand open parentheses cos theta plus isin theta close parentheses to the power of k equals sum from r equals 0 to k of open parentheses table row k row r end table close parentheses straight i to the power of left parenthesis k minus r right parenthesis end exponent cos to the power of r theta sin to the power of left parenthesis k minus r right parenthesis end exponent theta
  • STEP 3
    Use straight i squared equals negative 1 to simplify the expansion and group the real terms and the imaginary terms separately
  • STEP 4
    Equate the real parts of the expansion to cos kθ and equate the imaginary parts to sinkθ
  • STEP 5 (Depending on the question)
    Use sin squared theta plus cos squared theta equals 1 to write the identity in terms of sinθ only or cosθ depending on what the question asks
    • coskθ  can always be written as a function of just cosθ
    • sinkθ can be written as a function of just sinθ when k is odd
      • When k is even sinkθ will be a function of sinθ multiplied by a factor of cosθ

How do I write tankθ in terms of powers of tanθ?

  • STEP 1
    Find expressions for sinkθ and coskθ using the previous method
  • STEP 2
    Use the identity tan k theta equals fraction numerator sin k theta over denominator cos k theta end fraction
  • STEP 3
    Divide each term in the fraction by the highest power of cosθ to write each term using powers of tanθ and secθ
  • STEP 4 (Depending on the question)
    Write everything in terms of tanθ using the identity 1 plus tan squared theta equals sec squared theta

Exam Tip

  • You can use the substitutions c = cosθ and s = sinθ to shorten your working as long as you clearly state them and change back at the end of the proof

Worked example

Prove that sin 5 x identical to 16 sin to the power of 5 x minus 20 sin cubed x plus 5 sin x.

al-fm-1-2-3-apps-of-de-m-we-solution-1-fixed

Powers of Trig Functions

de Moivre’s theorem can be applied to prove identities such as sin to the power of 5 theta identical to 1 over 16 open parentheses 10 sin theta minus 5 sin 3 theta plus sin 5 theta close parentheses. This allows powers of a trig function to be written in terms of multiple angles which makes them easier to integrate.

How can I write coskθ and sinkθ in terms of e?

  • Recall straight e to the power of straight i theta end exponent equals cos theta plus isin theta and by de Moivre’s theorem straight e to the power of straight i k theta end exponent equals cos k theta plus isin k theta
  • It follows that straight e to the power of negative straight i k theta end exponent equals cos left parenthesis negative k theta right parenthesis plus isin left parenthesis negative k theta right parenthesis equals cos k theta minus isin k theta
  • You can derive expressions for sinkθ and coskθ using:
    • straight e to the power of straight i k theta end exponent plus straight e to the power of negative ikθ end exponent equals 2 cos k theta
      • cos k theta equals 1 half open parentheses straight e to the power of straight i k theta end exponent plus straight e to the power of negative straight i k theta end exponent close parentheses
    • straight e to the power of straight i k theta end exponent minus straight e to the power of negative straight i k theta end exponent equals 2 isin k theta
      • sin k theta equals fraction numerator 1 over denominator 2 straight i end fraction open parentheses straight e to the power of straight i k theta end exponent minus straight e to the power of negative straight i k theta end exponent close parentheses

How do I write powers of sinθ or cosθ in terms of sinkθ or coskθ?

  • STEP 1
    Write the trig term in terms of e
    • cos theta equals 1 half open parentheses straight e to the power of straight i theta end exponent plus straight e to the power of negative straight i theta end exponent close parentheses
    • sin theta equals fraction numerator 1 over denominator 2 straight i end fraction open parentheses straight e to the power of straight i theta end exponent minus straight e to the power of negative straight i theta end exponent close parentheses
  • STEP 2
    Use the binomial expansion to expand cos to the power of k theta equals 1 over 2 to the power of k open parentheses straight e to the power of straight i theta end exponent plus straight e to the power of negative straight i theta end exponent close parentheses to the power of k or sin to the power of k theta equals fraction numerator 1 over denominator 2 to the power of k straight i to the power of k end fraction open parentheses straight e to the power of straight i theta end exponent minus straight e to the power of negative straight i theta end exponent close parentheses to the power of k
    • Simplify ik to one of i, -1, -i or 1
  • STEP 3
    Due to symmetry you can pair terms up of the form A straight e to the power of straight i n theta end exponent and A straight e to the power of negative straight i n theta end exponent
    • Write as A open parentheses straight e to the power of straight i n theta end exponent plus straight e to the power of negative straight i n theta end exponent close parentheses or A open parentheses straight e to the power of straight i n theta end exponent minus straight e to the power of negative straight i n theta end exponent close parentheses
    • If k is even then there will be a term by itself as B straight e to the power of straight i m theta end exponent straight e to the power of negative straight i m theta end exponent equals B straight e to the power of 0 equals B
  • STEP 4
    Rewrite each pair in terms of cosnθ  or sinnθ
    •  A open parentheses straight e to the power of straight i n theta end exponent plus straight e to the power of negative straight i n theta end exponent close parentheses equals A left parenthesis 2 cos n theta right parenthesis
    • A open parentheses straight e to the power of straight i n theta end exponent minus straight e to the power of negative straight i n theta end exponent close parentheses equals A left parenthesis 2 isin n theta right parenthesis
  • STEP 5
    Simplify the expression – remember the 2k term!
    • coskθ can always be written as an expression using only terms of the form cosnθ
    • sinkθ can be written as an expression using only terms of the form:
      • sinnθ if k is odd
      • cosnθ if k is even

How do I write powers of tanθ in terms of sinkθ or coskθ?

  • Use tan to the power of k theta equals fraction numerator sin to the power of k theta over denominator cos to the power of k theta end fraction and use the previous steps
  • Note that the expression will be in terms of multiple angles of sin & cos and not tan

Worked example

Prove that sin to the power of 5 theta identical to 1 over 16 open parentheses 10 sin theta minus 5 sin 3 theta plus sin 5 theta close parentheses.

Uc~pYGCJ_al-fm-1-2-3-apps-of-de-m-we-solution-2

Trig Series

de Moivre’s theorem can be applied to find formulae for the sum of trigonometric series such as 

1 third cos theta plus 1 over 9 cos 3 theta plus 1 over 27 cos 5 theta plus... equals fraction numerator cos theta over denominator 5 minus 3 cos 2 theta end fraction

How can I find the sum of geometric series involving complex numbers?

  • The geometric series formulae work with complex numbers
    • S subscript n equals fraction numerator a open parentheses 1 minus r to the power of n close parentheses over denominator 1 minus r end fraction and S subscript infinity equals fraction numerator a over denominator 1 minus r end fraction (provided vertical line r vertical line less than 1)
  • Suppose w and z are two complex numbers then:
    • sum from k equals 1 to n of w z to the power of k minus 1 end exponent equals w plus w z plus w z squared plus... plus w z to the power of n minus 1 end exponent equals fraction numerator w left parenthesis 1 minus z to the power of n right parenthesis over denominator 1 minus z end fraction
    • sum from k equals 1 to infinity of w z to the power of k minus 1 end exponent equals w plus w z plus w z squared plus... equals fraction numerator w over denominator 1 minus z end fraction provided |z|<1
      • Compare these to the geometric series formulae with a=w and r=z

How can I find the sum of geometric series involving sinθ or cosθ?

  • Using de Moivre’s theorem: straight e to the power of straight i k theta end exponent equals cos k theta plus isin k theta
  • You can find coskθ and sinkθ by taking real and imaginary parts
    • cos k theta equals Re open parentheses straight e to the power of straight i k theta end exponent close parentheses
    • sin k theta equals Im open parentheses straight e to the power of straight i k theta end exponent close parentheses
  • Rewrite the series using eikθ to make it a geometric series
    • For example:
      • cos theta plus cos 4 theta plus cos 7 theta plus... equals Re open parentheses straight e to the power of straight i theta end exponent plus straight e to the power of 4 straight i theta end exponent plus straight e to the power of 7 straight i theta end exponent plus... close parentheses
      • sin theta plus sin 4 theta plus sin 7 theta plus... equals Im open parentheses straight e to the power of straight i theta end exponent plus straight e to the power of 4 straight i theta end exponent plus straight e to the power of 7 straight i theta end exponent plus... close parentheses
  • You can now use the formulae to find an expression for the sum
    • The series involving eikθ will be geometric so determine
      • whether it is finite or infinite
      • what is the value of (the first term) and (the common ratio)
  • Once you have used the formula the denominator will be of the form A minus B straight e to the power of straight i n theta end exponent
    • Multiply the numerator and denominator by A minus B straight e to the power of negative straight i n theta end exponent 
    • The denominator will become real A squared plus B squared minus 2 A B cos n theta
      • This is because straight e to the power of straight i n theta end exponent plus straight e to the power of negative straight i n theta end exponent equals 2 cos n theta
  • If your series involved sin terms then take the imaginary part of the sum
  • If your series involved cos terms then take the real part of the sum

Exam Tip

  • Exam questions normally lead you through this process
  • It is common for questions to let C equal the sum of the series with cos and let S equal the sum of the series with sin
  • You can then write C + iS which makes the trig terms becomes eikθ

Worked example

Prove that 1 third cos theta plus 1 over 9 cos 3 theta plus 1 over 27 cos 5 theta plus... equals fraction numerator cos theta over denominator 5 minus 3 cos 2 theta end fraction.

al-fm-1-2-3-apps-of-de-m-we-solution-3

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

  • Downloadable PDFs
  • Unlimited Revision Notes
  • Topic Questions
  • Past Papers
  • Model Answers
  • Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Dan

Author: Dan

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.