The $k$th derivative of $\sin^n x$ as $n \ \sin^{n - k} x$ times a polynomial in $\cos x$

Question

It seems the $k$th derivative of the function $x \mapsto \sin^n x$ can be expressed as $n \ \sin^{n - k} x$ times a polynomial with only even or only odd powers of $\cos x$ (depending on the parity of $k$), with polynomials in $n$ as coefficients, for natural $k$ and integer $n$ and complex $x$.

The highest term is $n^{k-1} \cos^k x$; the ranks of the other polynomial-in-$n$ coefficients for $\cos^{k-2j} x$ seem to be (maximally) $(k-1-j)$.

$$ \frac{\mathrm{d}^0}{\mathrm{d}x^0} \sin^n x = n \ \sin^{n - 0} x \ \left(n^{-1} \cos^0 x\right)\text{ for }n \neq 0 $$ $$ \frac{\mathrm{d}^1}{\mathrm{d}x^1} \sin^n x = n \ \sin^{n - 1} x \ \left(n^0 \cos x\right) $$ $$ \frac{\mathrm{d}^2}{\mathrm{d}x^2} \sin^n x = n \ \sin^{n - 2} x \ \left(n^1 \cos^2 x - 1\right) $$ $$ \frac{\mathrm{d}^3}{\mathrm{d}x^3} \sin^n x = n \ \sin^{n - 3} x \ \left(n^2 \cos^3 x + (-3 n + 2) \cos x\right) $$ $$ \frac{\mathrm{d}^4}{\mathrm{d}x^4} \sin^n x = n \ \sin^{n - 4} x \ \left(n^3 \cos^4 x + (-6 n^2 + 8 n - 4) \cos^2 x + (3 n - 2)\right) $$ $$ \frac{\mathrm{d}^5}{\mathrm{d}x^5} \sin^n x = n \ \sin^{n - 5} x $$ $$ \left(n^4 \cos^5 x + (-10 n^3 + 20 n^2 - 20 n + 8) \cos^3 x + (15 n^2 - 30 n + 16) \cos x\right) $$ $$ \frac{\mathrm{d}^6}{\mathrm{d}x^6} \sin^n x = n \ \sin^{n - 6} x $$ $$\left(n^5 \cos^6 x + (-15 n^4 + 40 n^3 - 60 n^2 + 48 n - 16) \cos^4 x + (45 n^3 - 150 n^2 + 196 n - 88) \cos^2 x + (-15 n^2 + 30 n - 16)\right) $$

Is there a general rule to determine the polynomial-in-$n$ coefficients?

Edit:

The OEIS sequence A133341 seems to determine the second polynomial-in-$n$ coefficients (when one alternates the signs, and discards the italic values):

• 1,
• 2,
• 1, 3,
• 3, 2, 4,
• 6, 8, 4, 5,
• 10, 20, 20, 8, 6,
• 15, 40, 60, 48, 16, 7,
• 21, 70, 140, 168, 112, 32, 8,
• 28, 112, 280, 448, 448, 256, 64, 9,
• 36, 168, 504, 1008, 1344, 1152, 576, 128, 10,
• 45, 240, 840, 2016, 3360, 3840, 2880, 1280, 256, …

Answer 1

This answer has since been written out in: https://arxiv.org/pdf/1911.01386.pdf . For the expressions with these polynomials, skip the section “Using the complex definitions and the binomial theorem”.

Determining $p_{k}(n)(\cos x)$ recursively

When, for natural $k$, and integer $n$, and complex $x$, $$ \frac{\mathrm{d}^k}{\mathrm{d}x^k} \sin^n x = \sin^{n - k} x \ p_{k}(n)(\cos x) $$ with $p_{k}(n)(\cos x)$ a polynomial in $\cos x$ with only terms with even powers of $\cos x$ when $k$ is even, and only terms with odd powers of $\cos x$ when $k$ is odd; with $n^k \cos^k x$ as the highest term;

and note that each coefficient of $p_{k}(n)(\cos x)$ includes the initial factor $n$ that I factored out in the question;

then $$ \frac{\mathrm{d}^{k+1}}{\mathrm{d}x^{k+1}} \sin^n x = \sin^{n - (k+1)} x \ \left( ((n - k) \cos x) \ p_{k}(n)(\cos x) + (\cos^2 x - 1) \frac{\mathrm{d}}{\mathrm{d}\cos x} p_{k}(n)(\cos x)\right) $$ with $\frac{\mathrm{d}}{\mathrm{d}\cos x} p_{k}(n)(\cos x)$ a polynomial in $\cos x$ with only terms with odd powers of $\cos x$ when $k$ is odd, and only terms with even powers of $\cos x$ when $k$ is even; with $k \ n \ \cos^{k-1} x$ as the highest term.

So $$ p_{k+1}(n)(\cos x) = ((n - k) \cos x) \ p_{k}(n)(\cos x) + (\cos^2 x - 1) \frac{\mathrm{d}}{\mathrm{d}\cos x} p_{k}(n)(\cos x) $$ is a polynomial in $\cos x$ with only terms with odd powers of $\cos x$ when $k$ is even, and only terms with even powers of $\cos x$ when $k$ is odd; with $((n - k)^1 \ n^k + k \ n) \ \cos^{k+1} x = n^{k+1} \ \cos^{k+1} x$ as the highest term.

Furthermore

For the same polynomials $p_{k}(n)(y)$ also holds $$ \frac{\mathrm{d}^k}{\mathrm{d}x^k} \cos^n x = (-1)^{k} \cos^{n - k} x \ p_{k}(n)(\sin x) $$ $$ \frac{\mathrm{d}^k}{\mathrm{d}x^k} \sinh^n x = \sinh^{n - k} x \ p_{k}(n)(\cosh x) $$