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I have two functions that I'm working on. The first is:

$$ \begin{align} \cos x &= (\cos 1)^3 \cos(3-x) \\ &{}+ 3 (\cos 1)^2 (\sin 1) \sin(3-x) \\ &{}- 3 (\cos 1) (\sin 1)^2 \cos(3-x) \\ &{}- (\sin 1)^3 \sin(3-x) \end{align} $$

...In other words, we can represent cosine as a series involving powers of cosines and sines. We can call the above function $f_3(x)$, since the maximum power of either sine or cosine is 3. In general, then, we can represent cosine as:

$$f_k(x) = \cos x = \sum_{i=0}^k c_{(k,i)} (\cos 1)^i (\sin 1)^{k-i}\cdot\left\{ \begin{array}{lr} \cos{(k-x)} & : i \text{ odd}\\ \sin{(k-x)} & : i \text{ even} \end{array} \right. $$

I've been trying to derive the coefficients ($c_{(k,i)} $) of this function. In general, I believe they are binomial coefficients, but I can't seem to figure out the pattern of signs (i.e. $+1$ or $-1$) for the coefficients. I'm wondering if anyone can get the formula for the coefficients of this function. I would greatly appreciate it.

Please note: $k$ is a natural.

Matt Groff
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  • @Blue: Hmm. I derived the formula by repeated sum formulas for sine and cosine. i.e. $\cos{(k-1+1)}=\cos{(k-1)}\cos{1}-\sin{(k-1)}\sin{1}$ and $\sin{(k-1+1)}=\sin{(k-1)}\cos{1}+\cos{(k-1)}\sin{1}$. I just repeatedly applied the formulas. Unfortunately, I can't seem to find the pattern of signs (i.e. positive/negative parity). Now I see my error... I'll fix the question. – Matt Groff Jan 23 '14 at 22:26

1 Answers1

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You have effectively decomposed $\cos x$ as $$\cos x = \cos( k - ( k - x ) ) = \cos k \cos(k-x) + \sin k \sin(k-x)$$ while expressing $\cos k$ and $\sin k$ using "multiple angle" expansions in terms of $\sin 1$ and $\cos 1$.

The binomial-like formulas for multiple angle formulas are well-known: $$\begin{align} \sin k\theta \quad&=\quad \sum_{j=0}^k {k \choose j} \; \cos^j\theta \; \sin^{k-j}\theta \;\cdot\; \sin\frac{(k-j)\pi}{2} \\ \cos k\theta \quad&=\quad \sum_{j=0}^k {k \choose j} \; \cos^j\theta \; \sin^{k-j}\theta \;\cdot\; \cos\frac{(k-j)\pi}{2} \end{align}$$ where those final factors ---$\sin\frac{(k-j)\pi}{2}$ and $\cos\frac{(k-j)\pi}{2}$--- are cleverly generating the $\pm 1$s (and term-killing $0$s) that seem to have stumped you.

The best way to understand the binomial-like nature of these formulas is to introduce DeMoivre's Formula from complex analysis: $$\cos k\theta + i \sin k \theta = \left(\; \cos\theta + i \sin\theta \;\right)^k$$ Expanding the right-hand side of DeMoivre's Formula using the Binomial Theorem, then equating real and imaginary parts of the left- and right-hand sides of the equation, gives the multiple-angle formulas.

Blue
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