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Does there exist a generalized formula for $\sin((2n-1)x)$?

I noticed that if, $\sin(1x)=t^1$

Then

$\sin(3x)=3t^1-4t^3$

$\sin(5x)=5t^1-20t^3+16t^5$

$\sin(7x)=7t^1-56t^3+112t^5-64t^7$

$\cdots$ They do appear to follow some relation but can we deduce and generalize the relation using Taylor polynomials?

Assuming, $\sin((2n-1)x)=a_1\sin^1(x)+a_2\sin^3(x)+a_3\sin^5(x)+a_4\sin^7(x)+\cdots?$

Paramanand Singh
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2 Answers2

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$$\cos(nx) + i \sin (nx) = (\cos (x) + i \sin (x))^n$$

$$\cos ((2n-1)x) + i \sin ((2n-1)x) = \sum_{k=0}^{2n-1} {2n-1 \choose k} (i \sin (x))^k (\cos (x))^{2n-1-k}$$

$$i \sin ((2n-1)x) = \sum_{j=1}^{n} {2n-1 \choose 2j-1} (i \sin (x))^{2j-1} \cos(x)^{2n-2j}$$

$i^{2j-1} = (-1)^j \cdot (-i)$, $\cos^2(x) =1-\sin^2(x)$ and making $\sin (x) = t$ gives us:

$$\sin ((2n-1)x) = -\sum_{j=1}^{n} {2n-1 \choose 2j-1} (-1)^j t^{2j-1} (1-t^2)^{n-j}$$

you may expand $(1-t^2)^{n-j}$ but first let's check some cases:

For $n=1$: $\sin (x) = - 1 \cdot (-1) t (1-t^2)^0 = t$

For $n=2$: $\sin (3x) = -[-3t(1-t^2) +t^3] = 3t-4t^3$

sounds good.

Now $$(1-t^2)^{n-j} = \sum_{p=0}^{n-j} {n-j \choose p}(-t^2)^{p}$$

$$\sin ((2n-1)x) = -\sum_{j=1}^{n} {2n-1 \choose 2j-1} (-1)^j t^{2j-1} \sum_{p=0}^{n-j} {n-j \choose p}(-t^2)^{p}$$

I do believe some work can be done from here, but I'm not able to continue because I'm not good with combinatorics. I'm not sure how to use Taylor series to make the coefficients more explicit.

EDIT: how could I forget, this is the Chebyshev polynomials of first kind for odd numbers: https://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html

EDIT 2: how about if we do $$\sin ((2n-1)x) = f(\sin (x))$$, then if $x = \arcsin (y) \implies f(y) = \sin((2n-1)\arcsin(y))$

and so, we only need to find the derivatives of $f$ at $0$.

$$f'(x) = \frac{(2n-1)\cos((2n-1)\arcsin(y))}{\sqrt{1-y^2}} \implies f'(0) =(2n-1)$$

so $f$ as a polynomial will always have a coefficient $2n-1$ on $t$. I think it is intuitive that the even derivatives are always zero, but it must be hard to find the next odd ones, as we have $\sqrt{1-y^2}$ on the denominator.

hellofriends
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These can also be written using regular polynomials in $\tan(x)$ with coefficients which are a subset of Pascal's triangle (https://oeis.org/A103327).

The polynomials are \begin{align} J_n(x)=\sum_{k=n}^{2n-1} {(-1)^{n-k}\binom{2n-1}{2(2n-k-1)} x^{2(k-n)}} \end{align}

And for $\sin((2n-1)x)$ we also have the general formula: \begin{align} \sin((2n-1)x)= \sin(x) \cos^{2(n-1)}(x)\;J_n\left(\tan(x)\right) \end{align}

I am not sure if there exists a reference with regards to these polynomials, but please let me know if you find one.

EXAMPLE:

Let $\sin(x)=t$ then we have of course $\cos(x)=\sqrt{1-t^2}$.

So we obtain the desired polynomials by substituting $\sin(x),\cos(x)$ in our general formula:

\begin{align} t (1-t^2)^{n-1}\; J_n\left(\frac{t}{\sqrt{1-t^2}}\right) \end{align}

This gives us :

\begin{align} n=1 \;: \;& t \\ n=2 \;: \;& 3 t - 4 t^3 \\ n=3 \;:\;& 5 t - 20 t^3 + 16 t^5 \\ n=4 \;:\;& 7 t - 56 t^3 + 112 t^5 - 64 t^7 \\ n=5 \;:\;& 9 t - 120 t^3 + 432 t^5 - 576 t^7 + 256 t^9 \\ n=6 \;:\;& 11 t - 220 t^3 + 1232 t^5 - 2816 t^7 + 2816 t^9 - 1024 t^{11}\\ ... \end{align}

Jakob
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