De Moivre trigonometric proofs
Master trigonometric proofs using De Moivre's theorem for Year 12 Specialist Mathematics in Queensland (QCAA, Unit 3). You expand \((\cos\theta+i\sin\theta)^n\) with the binomial theorem and equate it with \(\cos n\theta+i\sin n\theta\) to prove the multiple-angle identities.
You will learn to simplify the powers of \(i\), equate real and imaginary parts, and prove identities up to four theta such as \(\cos 3\theta=4\cos^3\theta-3\cos\theta\) — a key proof technique linking complex numbers and trigonometry in the Specialist course.
Theory
Trigonometric proofs using De Moivre's theorem derive the multiple-angle identities by algebra rather than by repeated angle-sum work. In Year 12 Specialist Mathematics (QCAA, Queensland, Unit 3) you expand \((\cos\theta+i\sin\theta)^n\) with the binomial theorem, set it equal to \(\cos n\theta+i\sin n\theta\), and equate the real and imaginary parts to prove identities such as \(\cos 3\theta=4\cos^3\theta-3\cos\theta\).
De Moivre's theorem states that for a positive integer \(n\), \((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\). Raising the unit complex number to a power multiplies its argument by \(n\); it does not raise the ratios to the power \(n\).
The right-hand side is already split into a real part \(\cos n\theta\) and an imaginary part \(\sin n\theta\). If the left-hand side is expanded with the binomial theorem and the powers of \(i\) are simplified using \(i^2=-1\), \(i^3=-i\) and \(i^4=1\), it too separates into a real part and an imaginary part.
Because two complex numbers are equal only when their real parts are equal and their imaginary parts are equal, you may equate parts: the real part of the expansion equals \(\cos n\theta\) and the imaginary part equals \(\sin n\theta\). This is the whole proof technique.
A final tidy-up with the Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\) rewrites each result in a single ratio, giving the standard multiple-angle identities up to \(4\theta\).
De Moivre's theorem is the starting point for every proof:
Expanding and equating parts for \(n=2\) and \(n=3\) gives:
Dividing the imaginary part by the real part gives the tangent form, and \(n=4\) reaches \(\cos 4\theta\):
How to prove a multiple-angle identity
- State De Moivre: write \((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\) for the value of \(n\) you need (\(2\), \(3\) or \(4\)).
- Expand the left side with the binomial theorem, using the Pascal coefficients (\(1,3,3,1\) for \(n=3\); \(1,4,6,4,1\) for \(n=4\)).
- Simplify the powers of \(i\): replace \(i^2=-1\), \(i^3=-i\), \(i^4=1\), then group the terms free of \(i\) (real) and the terms carrying \(i\) (imaginary).
- Equate parts: set the real part equal to \(\cos n\theta\) and the imaginary part equal to \(\sin n\theta\), then use \(\sin^2\theta=1-\cos^2\theta\) (or \(\cos^2\theta=1-\sin^2\theta\)) to finish in a single ratio.
Apply De Moivre, then expand the left side by the binomial theorem and use \(i^2=-1\):
| \((\cos\theta+i\sin\theta)^2\) | \(=\) | \(\cos 2\theta+i\sin 2\theta\) |
| \((\cos\theta+i\sin\theta)^2\) | \(=\) | \(\cos^2\theta+2i\cos\theta\sin\theta+i^2\sin^2\theta\) |
| \(=\) | \((\cos^2\theta-\sin^2\theta)+i(2\cos\theta\sin\theta)\) |
Equate the real parts and the imaginary parts of the two forms:
| \(\cos 2\theta\) | \(=\) | \(\cos^2\theta-\sin^2\theta\) |
| \(\sin 2\theta\) | \(=\) | \(2\cos\theta\sin\theta\) |
\(\cos 2\theta=\cos^2\theta-\sin^2\theta\) and \(\sin 2\theta=2\cos\theta\sin\theta\).
Expand \((\cos\theta+i\sin\theta)^3\) by the binomial theorem (coefficients \(1,3,3,1\)) and simplify the powers of \(i\):
| \((\cos\theta+i\sin\theta)^3\) | \(=\) | \(\cos^3\theta+3\cos^2\theta(i\sin\theta)+3\cos\theta(i\sin\theta)^2+(i\sin\theta)^3\) |
| \(=\) | \(\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\) | |
| \(=\) | \((\cos^3\theta-3\cos\theta\sin^2\theta)+i(3\cos^2\theta\sin\theta-\sin^3\theta)\) |
Equate real parts with \(\cos 3\theta\), then replace \(\sin^2\theta=1-\cos^2\theta\):
| \(\cos 3\theta\) | \(=\) | \(\cos^3\theta-3\cos\theta\sin^2\theta\) |
| \(=\) | \(\cos^3\theta-3\cos\theta(1-\cos^2\theta)\) | |
| \(=\) | \(\cos^3\theta-3\cos\theta+3\cos^3\theta\) | |
| \(=\) | \(4\cos^3\theta-3\cos\theta\) |
\(\cos 3\theta=4\cos^3\theta-3\cos\theta\).
The imaginary part of the expansion equals \(\sin 3\theta\):
| \(\sin 3\theta\) | \(=\) | \(3\cos^2\theta\sin\theta-\sin^3\theta\) |
Replace \(\cos^2\theta=1-\sin^2\theta\) and expand:
| \(\sin 3\theta\) | \(=\) | \(3(1-\sin^2\theta)\sin\theta-\sin^3\theta\) |
| \(=\) | \(3\sin\theta-3\sin^3\theta-\sin^3\theta\) | |
| \(=\) | \(3\sin\theta-4\sin^3\theta\) |
\(\sin 3\theta=3\sin\theta-4\sin^3\theta\).
Cube \(z+\dfrac{1}{z}\) and group the paired powers \(z^n+\dfrac{1}{z^n}\):
| \(\left(z+\dfrac{1}{z}\right)^3\) | \(=\) | \(z^3+3z+\dfrac{3}{z}+\dfrac{1}{z^3}\) |
| \(=\) | \(\left(z^3+\dfrac{1}{z^3}\right)+3\left(z+\dfrac{1}{z}\right)\) |
Replace each pair by \(2\cos n\theta\), noting \(\left(z+\dfrac{1}{z}\right)^3=(2\cos\theta)^3\):
| \((2\cos\theta)^3\) | \(=\) | \(2\cos 3\theta+3(2\cos\theta)\) |
| \(8\cos^3\theta\) | \(=\) | \(2\cos 3\theta+6\cos\theta\) |
| \(\cos^3\theta\) | \(=\) | \(\dfrac{\cos 3\theta+3\cos\theta}{4}\) |
\(\cos^3\theta=\dfrac{\cos 3\theta+3\cos\theta}{4}\).
Common pitfalls
Frequently asked questions
What is De Moivre's theorem?
For a positive integer \(n\), \((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\). Raising the unit complex number to a power multiplies its argument by \(n\).
How do you prove a multiple-angle identity with De Moivre's theorem?
Expand \((\cos\theta+i\sin\theta)^n\) with the binomial theorem, simplify the powers of \(i\), then equate the real part with \(\cos n\theta\) and the imaginary part with \(\sin n\theta\).
What does "equating parts" mean?
Two complex numbers are equal only when both their real parts and their imaginary parts are equal. Matching the real parts of the two forms, then the imaginary parts, gives two separate identities.
Why do the powers of i matter so much?
Because \(i^2=-1\), \(i^3=-i\) and \(i^4=1\), even powers of \(i\sin\theta\) land in the real part and odd powers in the imaginary part. A single slip mixes the two parts and spoils the identity.
Can De Moivre's theorem give tan 3θ?
Yes. Form \(\tan 3\theta=\dfrac{\sin 3\theta}{\cos 3\theta}\) from the imaginary and real parts, then divide numerator and denominator by \(\cos^3\theta\) to get \(\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}\).
How is this different from the double-angle method?
The double-angle method reaches the same identities by repeatedly applying the angle-sum formulae. The De Moivre method gets them in one binomial expansion followed by equating real and imaginary parts.