Seventy deadly sines

Consider the Maclaurin series $\sin(x + ax^3 + bx^5 + cx^7) \; = \; x + \big(a - \tfrac16)x^3 + \big(b - \tfrac12a + \tfrac{1}{120}\big)x^5 + \big(c - \tfrac12a^2 - \tfrac12b + \tfrac{1}{24}a - \tfrac{1}{5040}\big)x^7 + \cdots$ Suppose that $f_n(x)$ is the $n$ -fold composition of $\sin$ with itself, and suppose that $f_n$ has the Maclaurin expansion $f_n(x) \; = \; x + a_nx^3 + b_nx^5 + c_nx^7 + \cdots$ Then we deduce that $\begin{aligned} a_{n+1} & = \; a_n - \tfrac16 \\ b_{n+1} & = \; b_n - \tfrac12a_n + \tfrac{1}{120} \\ c_{n+1} & = \; c_n - \tfrac12a_n^2 - \tfrac12b_n + \tfrac{1}{24}a_n - \tfrac{1}{5040} \end{aligned}$ with $a_1 = -\tfrac16$ , $b_1 = \tfrac{1}{120}$ and $c_1 = -\tfrac{1}{5040}$ . Solving these recurrence relations in turn, we have $\begin{aligned} a_n & = \; -\tfrac16n \\ b_n & = \; \tfrac{1}{120}n(5n-4) \\ c_n & = \; -\tfrac{1}{15120}n(175n^2 - 336n + 164) \end{aligned}$ and we calculate $f_{70}^{(7)}(0) = 7! \times c_{70} \; = \; \boxed{-19463360}$ .

How to solve by sinning (WolframAlpha):

I didn't dare jump to 70 nested sines and seventh derivatives so I found smaller cases and looked for a pattern.

First, any even depth derivative with any number of nested sines will give 0 at x=0

First derivatives for 1,2,3... nested sines are 1, 1, 1, 1... ok, nice pattern.

Third derivatives go -1, -2, -3, -4... also nice and simple.

Fifth derivatives go 1, 12, 33, 64, 105... messier, but it turns out to be quadratic. $5n^{2}-4n$

I've got enough to guess at a pattern here. These are all polynomials with increasing degree. I'm guessing the next will fit a cubic.

Seventh derivatives go -1,-128, -731, -2160, -4765... It does indeed fit the cubic $-\frac{1}{3}(175n^{3}-336n^{2}+164n)$

Plug 70 into this to get the answer -19463360.

I've only shared this to show how one might approach a problem of this type if all else fails. My hat is off to those who used the Maclaurin series.

[Please don't Upvote this solution, it isn't worthy. Mark's is.]