How do I express powers like cosⁿθ and sinⁿθ in terms of cosines and sines of multiples of θ?

In one line

Put \(x=\cos\theta+i\sin\theta\), so \(2\cos\theta=x+\frac1x\) and \(2i\sin\theta=x-\frac1x\); raise to the power, expand, and pair terms into \(x^k\pm\frac1{x^k}\).

Explained simply

This is the reverse of the previous topic. Earlier we turned \(\cos4\theta\) into powers; now we turn \(\cos^4\theta\) into a sum of \(\cos4\theta,\cos2\theta\) and a constant. It is like breaking a big note into smaller coins. It is very useful in integration because \(\cos4\theta\) is easy to integrate but \(\cos^4\theta\) is not. Signal engineers use the same idea to find which harmonics (multiples of a base frequency) a power of a wave contains.

Technically, if \(x=e^{i\theta}\) then \(\frac1x=e^{-i\theta}\) and by De Moivre's theorem \(x^n+\frac1{x^n}=2\cos n\theta\), \(x^n-\frac1{x^n}=2i\sin n\theta\).

How it works

1. Write \((2\cos\theta)^n=\left(x+\frac1x\right)^n\) or \((2i\sin\theta)^n=\left(x-\frac1x\right)^n\). For mixed powers multiply both.
2. Expand by binomial theorem.
3. Pair the first and last terms, second and second-last, and so on, to form \(x^k\pm\frac1{x^k}\).
4. Replace each pair by \(2\cos k\theta\) or \(2i\sin k\theta\).
5. Simplify the powers of 2 and \(i\) on the left and divide.

Key formula

\(x+\frac1x=2\cos\theta,\ x-\frac1x=2i\sin\theta,\ x^n+\frac1{x^n}=2\cos n\theta,\ x^n-\frac1{x^n}=2i\sin n\theta\)

Here \(x=\cos\theta+i\sin\theta\) and \(n\) is a positive integer.

Quick example

Express \(\cos^4\theta\) in multiples of \(\theta\).

\((2\cos\theta)^4=\left(x+\frac1x\right)^4=x^4+4x^2+6+\frac4{x^2}+\frac1{x^4}=\left(x^4+\frac1{x^4}\right)+4\left(x^2+\frac1{x^2}\right)+6\).

So \(16\cos^4\theta=2\cos4\theta+8\cos2\theta+6\), giving \(\cos^4\theta=\dfrac{1}{8}(\cos4\theta+4\cos2\theta+3)\). Check at \(\theta=0\): \((1+4+3)/8=1\).

Similarly \((2i\sin\theta)^3=\left(x-\frac1x\right)^3=\left(x^3-\frac1{x^3}\right)-3\left(x-\frac1x\right)\) gives \(-8i\sin^3\theta=2i\sin3\theta-6i\sin\theta\), so \(\sin^3\theta=\dfrac{3\sin\theta-\sin3\theta}{4}\).

Common mistakes

• Forgetting that \((2i)^n\) brings a power of \(i\) on the left: \((2i)^3=-8i\), \((2i)^4=16\), \((2i)^6=-64\).
• Pairing terms with the wrong sign in \(\left(x-\frac1x\right)^n\) expansions, where signs alternate.
• Leaving the answer with \(x\) in it instead of converting back to cosines and sines.

How MU asks it

A standard 5/6-mark question: "Prove that \(\sin^5\theta=\dfrac1{16}(\sin5\theta-5\sin3\theta+10\sin\theta)\)" or "Express \(\cos^6\theta\) / \(\sin^4\theta\cos^2\theta\) in a series of cosines of multiples of \(\theta\)". For full marks, define \(x\), write the four basic relations, show the binomial expansion, the pairing and the final division.