Maclaurin series of $1- \cos^{2/3} x$ has all coefficients positive Experimenting with WA I noticed that the function $1- \cos^{\frac{2}{3}}x$ has the Maclaurin expansion with all coefficients positive ( works for any exponent in $[0, \frac{2}{3}]$). A trivial conclusion from this is $|\cos x|\le 1$, but it implies more than that, for instance see this.  Maybe   some ''natural'' proofs are available.  Thank you for your interest!
Note: an attempt  used a differential equation satisfied by the function. But the answer by @metamorphy just solved it the right way.
$\bf{Added:}$ Some comments about series with positive coefficients.
By $P$ we denote a series with positive coefficients ( no free term),

*

*If $a>0$ then  $\frac{1}{(1-P)^a} = 1+P$   (moreover, the positive expression on RHS is a polynomial in $a$ with positive coefficients


*If $0<a < 1$ then $(1-P)^a = 1-P$. Similarly the expressions for $a = \frac{t}{t+1}$ are positive in $t$.
2'. If $1-f= P$ then $1- f^{a} =P$ for any $0 < a < 1$, and similar with above.
$\bf{Added:}$ It turns out that the function $\cos^{2/3} x$ has a continued fraction (an $S$-fraction, from Stieltjes) that is "positive" ( similar to the continued fraction for $\tan x$). This is a stronger statement than the one before. Maybe there is some approach  using hypergeometric functions.
 A: This is not an answer but it is too long for a comment.
This is an interesting observation. I think that we just need to look at the coefficient of $x^4$; as soon as it is positive, all the next coefficients are positive too.
There are amazing patterns.
Let $a=\frac {2 \times 10^k+1}{3 \times 10^k}$. The coefficient of $x^4$ in the expansion of $\big[1-\cos^a(x)\big]$ write
$$c_4(k)=-\frac {\alpha_k}{24\times 10^{2k}}$$ and the $\alpha_k$ form the sequence
$$\{1,7,67,667,6667,66667,666667,\cdots\}$$
Let $b=\frac {2 \times 10^k-1}{3 \times 10^k}$. The coefficient of $x^4$ in the expansion of $\big[1-\cos^b(x)\big]$ write
$$c_4(k)=\frac {\beta_k}{72\times 10^{2k}}$$ and the $\beta_k$ form the sequence
$$\{1,19,199,1999,19999,199999,1999999,\cdots\}$$
A: Start with the positive series
$$\csc x = \frac{1}{x} + \frac{1}{6} x  + \frac{7}{360} x^3  + \frac{31}{15120} x^5  + \cdots $$
Take the square and obtain the positive series
$$\csc^2 x =\frac{1}{x^2}  + \frac{1}{3}  + \frac{1}{15} x^2  + \frac{2}{189} x^4 + \cdots $$
Now take the derivative and get the  series
$$-  \frac{2\cos x}{\sin^3 x} = -\frac{2}{x^3} + \frac{2}{15} x + \frac{8}{189} x^3 + \cdots $$
where all the coefficients after the first are positive. We conclude that the series
$$ 1 - \frac{x^3 \cos x}{ \sin ^3 x}=\frac{1}{15}x^4  + \frac{4}{189}x^6 + \frac{1}{225}x^8  + \cdots $$
is positive. Now, since  $(1-t)^{-1/3}$ is a positive series in $t$, if we substitute for $t$ a positive series in $x$ without free term we get again a positive series in $x$. We conclude that
$$\frac{\sin x}{x \cos^{1/3} x}= 1 + \frac{1}{45}x^4  + \frac{4}{567}x^6 + \frac{1}{405} x^8+ \cdots $$
is a positive series. Now multiply by $\frac{2}{3} x$ and integrate and get the positive $1- \cos^{2/3} x$  ( all coefficients are, except perhaps the free term, which turns out to be $0$).
$\bf{Note:}$ In fact we can prove that the series for
$$\sin x - \frac{x^3 \cos x}{ \sin^2 x}$$
is positive. For this, we start with the expansion
$$\csc x= x^{-1} + \frac{1}{6} x + \frac{7}{360}x^3 + \frac{31}{15120} x^5 + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^{n+1} 2(2^{2n-1} -1) B_{2n}}{(2n)!}x^{2n-1}$$
Taking the derivative with respect to $x$ we get the expansion of $\frac{\cos x}{\sin^2 x}$. From here we conclude that $\sin x- \frac{x^3 \cos x}{\sin^2 x}$ is positive. Multiplying by the positive $\csc x$ we conclude that $1 - \frac{x^3 \cos x}{\sin^3 x}$ is positive.
A: (A proof, not very "natural" though.) $f(x)=1-\cos^{2/3}x$ satisfies $xf''(x)=g(x)f'(x)$, where $$g(x)=x\cot x+\frac{x}3\tan x=1+\sum_{n=1}^\infty g_n x^n$$ with $g_n=0$ for odd $n$, and $3g_{2n}=(-1)^n 2^{2n}(4-2^{2n})B_{2n}/(2n)!\geqslant 0$ using Bernoulli numbers (and the alternating-sign property of these).
Now $f'(x)=\sum_{n=1}^\infty f_n x^n$ implies $(n-1)f_n=\sum_{k=1}^{n-1}g_{n-k}f_k$, giving $f_n\geqslant 0$ by induction.
A: Some thoughts:
Let $f(x) := \cos^{2/3} x$.
We have
$$f'(x) = \frac23 \cos^{-1/3} x\, (-\sin x) = -\frac23 f(x)\tan x $$
and
$$f''(x) = -\frac23 f'(x) \tan x - \frac23 f(x) (\tan x)' = - \frac23 f(x) - \frac29 f(x)\tan^2 x .$$
Conjecture 1: For $n = 1, 2, \cdots$,
$$f^{(2n - 1)}(x) = f(x) \sum_{k=1}^n c_{2n-1, k} \tan^{2k-1} x$$
and
$$f^{(2n)}(x) = f(x) c_{2n, 0} + f(x) \sum_{k=1}^n c_{2n, k} \tan^{2k} x$$
where $c_{2n-1, k} \le 0$ for all $n\ge 1,\, 1\le k \le n$,
and $c_{2n, k} \le 0$ for all $n\ge 1,\, 0 \le k \le n$.
I think we can use Mathematical Induction or Strong Induction to prove it.
