Expansions of sin(nx) and cos(nx)

If you have gone through double-angle formula or triple-angle formula, you must have learned how to express trigonometric functions of \(2\theta\) and \(3\theta\) in terms of \(\theta\) only. In this wiki, we'll generalize the expansions of various trigonometric functions.

We have the following general formulas:

• \({\sin(n\theta) = \displaystyle \sum_{r=0 \atop 2r+1 \leq n} (-1)^r \binom{n}{2r+1} \cos^{n-2r-1}(\theta) \sin^{2r+1}(\theta)}\)

• \({\cos(n\theta) = \displaystyle \sum_{r=0 \atop 2r \leq n} (-1)^r \binom{n}{2r} \cos^{n-2r}(\theta) \sin^{2r}(\theta)}\)

• \({\tan(n\theta) = \frac{\displaystyle \sum_{r=0 \atop 2r+1 \leq n} (-1)^r \binom{n}{2r+1} \tan^{2r+1}(\theta)}{\displaystyle \sum_{r=0 \atop 2r \leq n} (-1)^r \binom{n}{2r} \tan^{2r}(\theta)}}\)

Display The Proof

Let's invoke De Moivre's theorem here. We have

\[\cos(n\theta) + i\sin(n\theta) = \big(\cos(\theta) + i\sin(\theta) \big)^n.\]

Since \(n\) is a positive integer, the binomial theorem-N choose K holds for \(\big(\cos(\theta) + i\sin(\theta) \big)^n\).

Hence, by expanding, we have

\[\big(\cos(\theta) + i\sin(\theta) \big)^n= \cos^n(\theta) + n\cos^{n-1}(\theta) \cdot i \sin(\theta) + \frac{n(n-1)}{1\cdot 2} \cos^{n-2}(\theta) \cdot i^2 \sin^2(\theta)+ \cdots.\]

Since \(i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, \ldots,\) we have

\[\begin{align} \big(\cos(\theta) + i\sin(\theta) \big)^n =&\left[\cos^n(\theta) - \dfrac{n(n-1)}{1\cdot 2} \cos^{n-2}(\theta) \cdot \sin^2(\theta) + \dfrac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4} \cos^{n-4}(\theta) \cdot \sin^4(\theta) + \cdots\right] \\ &+ i \left[ n\cos^{n-1}(\theta) \cdot \sin(\theta) -\dfrac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3 } \cos^{n-3}(\theta) \cdot \sin^3(\theta) + \cdots \right]. \end{align}\]

By equating the real and imaginary parts, we obtain

\[\begin{align} \cos(n\theta) &= \cos^n(\theta) - \dfrac{n(n-1)}{1\cdot 2} \cos^{n-2}(\theta) \cdot \sin^2(\theta) + \dfrac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4} \cos^{n-4}(\theta) \cdot \sin^4(\theta) + \cdots \\ \sin(n\theta) &= n\cos^{n-1}(\theta) \cdot \sin(\theta) -\dfrac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3 } \cos^{n-3}(\theta) \cdot \sin^3(\theta) + \cdots, \end{align} \]

where the terms of both the above series are alternately positive and negative. Also, each series continues till one of the factors in the numerator is zero and then ceases. Hence, proved.

To prove for \(\tan(n\theta)\), observe that

\[\tan(n\theta) = \dfrac{\sin(n\theta)}{\cos(n\theta)} = \dfrac{\displaystyle \sum_{r=0 \atop 2r+1 \leq n} (-1)^r \binom{n}{2r+1} \cos^{n-2r-1}(\theta) \sin^{2r+1}(\theta)}{\displaystyle \sum_{r=0 \atop 2r \leq n} (-1)^r \binom{n}{2r} \cos^{n-2r}(\theta) \sin^{2r}(\theta)}.\]

Divide the numerator and the denominator by \(\cos^n(\theta)\) to obtain

\[{\tan(n\theta) = \frac{\displaystyle \sum_{r=0 \atop 2r+1 \leq n} (-1)^r \binom{n}{2r+1} \tan^{2r+1}(\theta)}{\displaystyle \sum_{r=0 \atop 2r \leq n} (-1)^r \binom{n}{2r} \tan^{2r}(\theta)}}.\]

Hence, our proof is complete. \(_\square\)

For angles \(\theta_1, \theta_2, \theta_3, \ldots\), we have

• \({\sin(\theta_1 + \theta_2 + \theta_3 + \cdots) = \cos(\theta_1) \cdot \cos(\theta_2) \cdot \cos(\theta_3) \dotsm [s_1 - s_3 + s_5 - s_7 + \cdots]}\)
• \({ \cos(\theta_1 + \theta_2 + \theta_3 + \cdots) = \cos(\theta_1) \cdot \cos(\theta_2) \cdot \cos(\theta_3) \dotsm [1 - s_2 + s_4 - s_6 + \cdots]}\)
• \({\tan(\theta_1 + \theta_2 + \theta_3 + \cdots) = \dfrac{s_1 - s_3 + s_5 - s_7 + \cdots}{1 - s_2 + s_4 - s_6 + \cdots}},\)

where \({s_n = \displaystyle \sum_\text{cyclic} \tan(\theta_1) \tan(\theta_2) \tan(\theta_3) \dotsm \tan(\theta_n)}.\)

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We'll show here, without using any form of Taylor's series, the expansion of \(\sin(\theta), \cos(\theta), \tan(\theta)\) in terms of \(\theta\) for small \(\theta\).

Here are the generalized formulaes:

• \({\sin(\theta) = \displaystyle \sum_{r=0}^\infty (-1)^r \dfrac{\theta^{2r+1}}{(2r+1)!}}\)
• \({\cos(\theta) = \displaystyle \sum_{r=0}^\infty (-1)^r \dfrac{\theta^{2r}}{(2r)!}}\)
• \({\tan(\theta) = \theta + \dfrac{\theta^3}{3} + \dfrac{2\theta^5}{15} + \cdots}.\)

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We have the following general formulas:

If \(n\) is even,

• \({\cos^n(\theta) = \dfrac{1}{2^{n-1}} \left[ \displaystyle \sum_{r=0 \atop 2r < n} \binom{n}{2r} \cos \big( (n-2r)\theta \big) \right] + \dfrac{1}{2^n} \displaystyle\binom{n}{n/2}}\)
• \({\sin^n(\theta) = \dfrac{(-1)^{n/2}}{2^{n-1}} \left[ \displaystyle \sum_{r=0 \atop 2r < n} (-1)^r \binom{n}{2r} \cos \big( (n-2r)\theta \big)\right] + \dfrac{1}{2^n} \displaystyle\binom{n}{n/2} }.\)

Display the Proof

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If \(n\) is odd,

• \({\cos^n(\theta) = \dfrac{1}{2^{n-1}} \left[ \displaystyle \sum_{r=0 \atop 2r < n} \binom{n}{2r} \cos \big( (n-2r)\theta \big) \right] }\)
• \({\sin^n(\theta) = \dfrac{(-1)^{n/2}}{2^{n-1}} \left[ \displaystyle \sum_{r=0 \atop 2r < n} (-1)^r \binom{n}{2r} \sin \big( (n-2r)\theta \big) \right] }.\)

Display the Proof

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