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),
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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
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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.
Best Answer
(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.