# 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.

#### Contents

- Expansion of \(\sin(n\theta), \cos(n\theta),\) and \(\tan(n\theta)\)
- Generalized Version of the above Expansion
- Expanding \(\sin(\theta), \cos(\theta), \tan(\theta)\) in terms of \(\theta\) for small \(\theta\)
- Expanding \(\sin^n(\theta), \cos^n(\theta),\) and \(\tan^n(\theta)\) in terms of \(k\theta\)

## Expansion of \(\sin(n\theta), \cos(n\theta),\) and \(\tan(n\theta)\)

We have the following general formulas:

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

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

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

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

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

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

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

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

## Generalized Version of the above Expansion

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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## Expanding \(\sin(\theta), \cos(\theta), \tan(\theta)\) in terms of \(\theta\) for small \(\theta\)

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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## Expanding \(\sin^n(\theta), \cos^n(\theta),\) and \(\tan^n(\theta)\) in terms of \(k\theta\)

We have the following general formulas:

If \(n\) is even,

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

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If \(n\) is odd,

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

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**Cite as:**Expansions of sin(nx) and cos(nx).

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/expansions-of-certain-trigonometric-functions/