Computing the value of $\zeta(6)$ and $\zeta(8)$ I know the closed form formula for calculating the values of Zeta function at even integers. I was able to derive it from the coefficients of the power series of $-\dfrac{\pi x}{2}\cot(\pi x)$. Now, I am looking at the way with which Euler was able to derive the values of $\zeta(2n)$.
I have seen this "trick" in Brilliant:
So I have:
$$\dfrac{\sin(x)}{x}=1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\dfrac{x^6}{7!}+\dfrac{x^8}{9!}...\tag{1}$$
$$\dfrac{\sin(x)}{x}=\left(1-\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{4\pi^2}\right)\left(1-\dfrac{x^2}{9\pi^2}\right)\left(1-\dfrac{x^2}{16\pi^2}\right)...\tag{2}$$
$$\dfrac{\sin(ix)}{ix}=1-\dfrac{(ix)^2}{3!}+\dfrac{(ix)^4}{5!}-\dfrac{(ix)^6}{7!}...\tag{3}$$
$$\dfrac{\sin(ix)}{ix}=1+\dfrac{x^2}{3!}+\dfrac{x^4}{5!}+\dfrac{x^6}{7!}...\tag{4}$$
$$\dfrac{\sin(ix)}{ix}=\left(1-\dfrac{(ix)^2}{\pi^2}\right)\left(1-\dfrac{(ix)^2}{4\pi^2}\right)\left(1-\dfrac{(ix)^2}{9\pi^2}\right)\left(1-\dfrac{(ix)^2}{16\pi^2}\right)...\tag{5}$$
$$\dfrac{\sin(ix)}{ix}=\left(1+\dfrac{x^2}{\pi^2}\right)\left(1+\dfrac{x^2}{4\pi^2}\right)\left(1+\dfrac{x^2}{9\pi^2}\right)\left(1+\dfrac{x^2}{16\pi^2}\right)...\tag{6}$$
Multiply $(6)$ and $(2)$, I have:
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=1+\left(\dfrac{x^2}{3!}-\dfrac{x^2}{3!}\right)+\left(\dfrac{x^4}{5!}+\dfrac{x^4}{5!}-\dfrac{x^4}{(3!)^2}\right)+...\tag{7}$$
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=\left(1+\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{\pi^2}\right)\left(1+\dfrac{x^2}{4\pi^2}\right)\left(1-\dfrac{x^2}{4\pi^2}\right)\left(1+\dfrac{x^2}{9\pi^2}\right)\left(1-\dfrac{x^2}{9\pi^2}\right)...\tag{8}$$
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=\left(1-\dfrac{x^4}{\pi^4}\right)\left(1-\dfrac{x^4}{16\pi^4}\right)\left(1-\dfrac{x^4}{81\pi^4}\right)...\tag{9}$$
Comparing the coeffcients of $x^4$ in the infinite Maclaurin series and the infinite product, we will arrive at $\zeta(4)=\dfrac{\pi^4}{90}$.
Now that I wish to continue this process to see if I can get $\zeta(6)$, $\zeta(8)$, $\zeta(10)$, I don't know how to continue. The trick above works fine for $\zeta(4)$, but what about higher $\zeta(2n)$.
I look around and find this interesting post of user17762:
In his answer, he wrote the function on the LHS as $\dfrac{i\sin(z)\sin(\dfrac{z}{i})}{z^2}$. A little bit of manipulation shows that it is exactly like $\dfrac{\sin(x)\sin(ix)}{ix^2}$  or $\dfrac{\sin(x)\sinh(x)}{x^2}$
I have tried to multiply again $\dfrac{\sin(x)\sinh(x)\sin(x/i^2)}{x^2(\frac{x}{i^2})}=\dfrac{\sin^2(x)\sinh(x)}{x^3}$, using the hint found in the answer of the member above. Then I use Wolfram to check the series expansion of this function but the third term of $x^6$ is $\dfrac{5}{3024}$. I thought that I should a find a function where its coefficient of $x^6$ should be $\dfrac{1}{945}$. I am not sure if my reasoning is problematic.
Is there an algorithm to find the function on the LHS?
 A: The generalisation you are looking for is given by the functions
$$f_n \colon \mathbb{R} \to \mathbb{R}, \, f_n (x) = \prod \limits_{k=0}^{n-1} \operatorname{sinc}\left(\mathrm{e}^{\mathrm{i} \pi \frac{k}{n}} x\right) \, , $$
for $n \in \mathbb{N}$ ($\operatorname{sinc}(z) = \frac{\sin(z)}{z}$ with $\operatorname{sinc}(0) = 1$). Since the product representation of $\operatorname{sinc}$ converges absolutely, we can rearrange factors to obtain
$$ f_n(x) = \prod \limits_{k=0}^{n-1} \prod \limits_{l=1}^\infty \left(1 - \mathrm{e}^{2 \pi \mathrm{i} \frac{k}{n}} \frac{x^2}{\pi^2 l^2} \right) = \prod \limits_{l=1}^\infty \prod \limits_{k=0}^{n-1} \left(1 - \mathrm{e}^{2 \pi \mathrm{i} \frac{k}{n}} \frac{x^2}{\pi^2 l^2} \right) = \prod \limits_{l=1}^\infty \left(1 - \left(\frac{x}{\pi l}\right)^{2n} \right)$$
for $x \in \mathbb{R}$ and $n \in \mathbb{N}$ (see this question for a simple proof of the final step). This representation confirms that these functions are real at real arguments. Clearly, the coefficient in front of $x^{2n}$ in the series expansion of $f_n$ is $[x^{2n}] f_n (x) = - \frac{\zeta(2n)}{\pi^{2n}}$ for $n \in \mathbb{N}$. Now simply multiply the power series representations, i.e.
$$ f_n (x) = \prod \limits_{k=0}^{n-1}  \sum \limits_{l_k = 0}^\infty \frac{(-1)^{l_k} \mathrm{e}^{2 \pi \frac{k l_k}{n} \mathrm{i}} x^{2l_k}}{(2l_k+1)!} \, ,$$
and compute the coefficient this way to obtain the zeta values by comparison. As you can easily check using CAS, this works for every $n \in \mathbb{N}$ ($[x^6]f_3(x) = - \frac{1}{945}$ or $[x^8]f_4(x) = - \frac{1}{9450}$, for example), but I have not yet managed to prove $[x^{2n}] f_n(x) = \frac{(-1)^{n} \mathrm{B}_{2n} 2^{2n-1}}{(2n)!}$ analytically.
