Unexpected cancellation in the computation of $\zeta(8)$: is there a structural reason?

Question

I am studying an Euler-type recursive method for computing even values of the Riemann zeta function.

The basic idea is to compare the Taylor expansion of $\sin z$ with Euler’s infinite product expansion, and to organize lattice sums using symmetry. By refining this approach, one obtains recursive relations between even zeta values.

For example, the method yields $$\zeta(4) = \zeta(2)^2 - \frac{2\pi^4}{5!},$$ where the term $2\pi^4/5!$ can be identified directly from the Taylor expansion of $\sin x$.

Derivation of $\zeta(4)=\zeta(2)^2-\dfrac{2\pi^4}{5!}$

Start from Euler's product for $\sin z$: $$ \frac{\sin z}{z}=\prod_{n=1}^{\infty}\left(1-\frac{z^2}{(n\pi)^2}\right). $$ Write the product as a power series $$ \prod_{n=1}^{\infty}\left(1-\frac{z^2}{(n\pi)^2}\right) =1-Az^2+Bz^4+O(z^6), $$ where $$ A=\sum_{n\ge1}\frac{1}{(n\pi)^2},\qquad B=\sum_{1\le m<n}\frac{1}{(m\pi)^2(n\pi)^2}. $$ Hence $$ \sin z =z\left(1-Az^2+Bz^4+O(z^6)\right) = z-Az^3+Bz^5+O(z^7). $$

On the other hand, the Taylor series is $$ \sin z = z-\frac{z^3}{3!}+\frac{z^5}{5!}+O(z^7). $$

Comparing coefficients gives $$ A=\frac{1}{3!},\qquad B=\frac{1}{5!}. $$ From $A=\sum_{n\ge1}\frac{1}{(n\pi)^2}$) we recover $$ \zeta(2)=\sum_{n\ge1}\frac{1}{n^2}=\pi^2 A=\frac{\pi^2}{6}. $$ From $B=\sum_{m<n}\frac{1}{(m\pi)^2(n\pi)^2}$ we get $$ \sum_{m<n}\frac{1}{m^2n^2}=\pi^4 B=\frac{\pi^4}{5!}. $$

Now decompose the square (lattice-sum viewpoint): $$ \zeta(2)^2=\left(\sum_{n\ge1}\frac1{n^2}\right)^2 =\sum_{m,n\ge1}\frac{1}{m^2n^2} =\sum_{n\ge1}\frac{1}{n^4} +2\sum_{m<n}\frac{1}{m^2n^2}. $$ That is, $$ \zeta(2)^2=\zeta(4)+2\sum_{m<n}\frac{1}{m^2n^2}. $$ Substituting $\sum_{m<n}\frac{1}{m^2n^2}=\frac{\pi^4}{5!}$ yields $$ \boxed{\ \zeta(4)=\zeta(2)^2-\frac{2\pi^4}{5!}\ }. $$ In particular, using $\zeta(2)=\pi^2/6$ gives $\zeta(4)=\pi^4/90 $.

Using the same recursive idea, I applied this method further to compute $$ \zeta(8) = \pi^8 \left[ \frac{1}{12} \left( \frac{1}{15120} + 24\left(\frac{1}{7!\cdot 6}-\frac{4}{9!}\right) + \frac{6}{90\cdot 90} + \frac{4}{945\cdot 6} - \frac{1}{1296} \right) \right]. $$

During this computation, the following unexpected cancellation appeared naturally:

$$ \frac{1}{15120} + \frac{4}{945\cdot 6} - \frac{1}{1296} = 0. $$

Each term in this identity originates from a different component of the recursive construction: lower-order zeta values, Taylor coefficients of $\sin x$, and symmetry-reduced lattice sums.

My question:

Is there a conceptual or structural explanation for the cancellation identity that appears naturally in this recursive Euler-type computation of even zeta values? Any insight or references would be appreciated.