Trivial (?) product/series expansions for sine and cosine In an old paper of Glaisher, I find the following formulas:
$$\dfrac{\sin(\pi x)}{\pi x}=1-\dfrac{x^2}{1^2}-\dfrac{x^2(1^2-x^2)}{(1.2)^2}-\dfrac{x^2(1^2-x^2)(2^2-x^2)}{(1.2.3)^2}-\cdots$$
$$\cos(\pi x/2)=1-x^2-\dfrac{x^2(1^2-x^2)}{(1.3)^2}-\dfrac{x^2(1^2-x^2)(3^2-x^2)}{(1.3.5)^2}-\cdots$$
These are trivial since the $n$th partial sum is equal to the $n$th partial product of
the product formulas for the sine and cosine
But he also gives
$$\sin(\pi x/2)=x-\dfrac{x(x^2-1^2)}{3!}+\dfrac{x(x^2-1^2)(x^2-3^2)}{5!}-\cdots$$
$$\cos(\pi x/2)=1-\dfrac{x^2}{2!}+\dfrac{x^2(x^2-2^2)}{4!}-\dfrac{x^2(x^2-2^2)(x^2-4^2)}{6!}+\cdots$$
$$\dfrac{\sin(\pi x/3)}{\sqrt{3}/2}=x-\dfrac{x(x^2-1^2)}{3!}+\dfrac{x(x^2-1^2)(x^2-2^2)}{5!}-\cdots$$
$$\cos(\pi x/3)=1-\dfrac{x^2}{2!}+\dfrac{x^2(x^2-1^2)}{4!}-\dfrac{x^2(x^2-1^2)(x^2-2^2)}{6!}+\cdots$$
Apparently he considers them trivial. If they are, please explain and feel free to
downvote me.

You can derive at least some of these formulas from expansions of $\sin x\theta$ and $\cos x\theta$ as Taylor series in $\sin\theta$,
$$\sin x\theta=x\sin \theta-\frac{x(x^2-1)}{3!}\sin^3\theta+\frac{x(x^2-1)(x^2-3^2)}{5!}\sin^5\theta+-\cdots$$
$$\cos x\theta=1-\frac{x^2}{2!}\sin^2 \theta+\frac{x^2(x^2-2^2)}{4!}\sin^4\theta-\frac{x^2(x^2-2^2)(x^2-4^2)}{6!}\sin^6\theta+-\cdots$$
(For a worked out proof, see this MSE post.)
Substituting $\theta=\pi/2$ gives the $\sin(\pi x/2)$ and $\cos(\pi x/2)$ series. Substituting $\theta=\pi/3$ gives slightly different series than in the OP...
