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Maybe this is a well-know question, anyway I haven't found an exact duplicate.

It is possible to express $\cos (nx)$ as a polynomial of degree $n$ in $\cos(x)$. As stated in this answer, it is possible to prove it starting from $\cos(nx)$ by repeatedly using

$$ \begin{align*} \cos(a\pm b) &= \cos a\cos b \mp \sin a\sin b\\ \sin(a\pm b) &= \sin a\cos b \pm \cos a\sin b\\ \sin^2(r) &= 1-\cos^2(r). \end{align*} $$

Now I would like to obtain $\sin(nx)$ as a polynomial of degree $n$ in $\sin(x)$. I started from the simplest case:

$$\sin(2x) = \sin x \cos x + \cos x \sin x = 2 \sin x \cos x$$

All I could do was to use $\cos(2x) = 1 - 2 \sin^2 x$, but then I would get

$$\sin(2x) = 2 \sin x \left[ 1 - 2\sin^2 \left( \frac{x}{2} \right) \right]$$

This is a polynomial with $\sin$, but not always $\sin(x)$. Moreover, this answer shows that $\sin(2x)$ can't be written as a polynomial in $\sin x$.

So, for $n \neq 2$, is it possible to express $\sin(nx)$ in terms of $\sin x$ or it is always impossible?

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BowPark
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1 Answers1

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So as far as I know the easiest way to figure out a way to express $\sin(nx)$ would be to use De Moivre's Theorem, which basically states that

$$\cos(nx)+i\sin(nx)=\left[ \cos(x)+i\sin(x) \right]^n$$

If you're unfamiliar with complex numbers all you need to know for now is that $i^2=-1$.

So lets try using that to compute $\sin(3x)$ and $\sin(4x)$:

$\cos(3x)+i\sin(3x)=\cos^3(x)+3i\cos^2(x)\sin(x)-3\cos(x)\sin^2(x)-i\sin^3(x)$

(it was $\cos^3(x)-3\cos(x)\sin^2(x))+i(3\cos^2(x)\sin(x)-\sin^3(x)$)

from here it is easy to see that $\sin(3x)=3\cos^2(x)\sin(x)-\sin^3(x)$

the main thing here is that the exponents on the $\cos$ is always even in this case which would allow us to us the identity $\cos^2(x)=1-\sin^2(x)$. Substituting that back in our equation we get:

$$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$

Now we can repeat the exact same process with $\sin(4x)$ to get:

$\cos(4x) + i\sin(4x)=\cos^4(x)+4i\cos^3(x)\sin(x)-6\cos^2(x)\sin^2(x)-4i\cos(x)\sin^3(x)+\sin^4(x).$

From this we have:

$$\sin(4x)=4\cos^3(x)\sin(x)-4\cos(x)\sin^3(x)$$

Notice that the exponents on the $\cos$ is odd and so we can't express $\sin(4x)$ as a polynomial in $\sin(x)$. With a little bit of thought you can see how this would continue. For odd $n$ you can indeed write $\sin(nx)$ as a polynomial.

Note also that the same thing happens with $\cos(nx)$ but instead when $n$ is even then you can write it as such polynomial.

BowPark
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H_Hassan
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  • thank you! Just a clarification: with $\cos (nx)$, anyway, both for $n$ even and for $n$ odd it is possible to create the polynomial, while for $\sin (nx)$ the polynomial can be created just when $n$ is odd. Is it right? – BowPark Jun 15 '16 at 10:28
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    For $\cos(nx)$ you'd find that the exponents for $\cos(x)$ would be odd in the expansion whenever $n$ is odd, and so you wouldn't be able to express it as a polynomial in $\sin(x)$. So they alternate. Whenever $n$ is odd, then you can express $\sin(nx)$ as a polynomial in $\sin(x)$ and whenever $n$ is even you can express $\cos(nx)$ as a polynomial in $\sin(x)$ – H_Hassan Jun 15 '16 at 15:21
  • Oh, sure, I haven't noticed this. But what I would like to point out is: while $\cos(nx)$ is always (both for $n$ even and odd) expandable as a polynomial in the variable $p = \cos (x)$, $\sin (nx)$ is expandable as a polynomial in the variable $q = \sin (x)$ only when $n$ is odd. So, in this sense, the sine function is not perfectly the dual of the cosine (and this sounds weird to me). Is it true? – BowPark Jun 15 '16 at 15:36