<s>[INST] By using Wallis’s product, prove $\zeta(2)=\frac{\pi^2}{6}$ $\sin x = x\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{(2\pi)^2}\right)\cdots$  
This equal is used in both the solution of  Basel’s problem and the proof of Wallis’s product.
So I wonder if I can prove one by using the other.
Help me solve this problem.
Or if you are sure this assumption is wrong, tell me the reason.   
I’m sorry if I put wrong tugs.
 [/INST] You get the $\,$ Wallis product $\,$, if you set $\,\displaystyle x=\frac{\pi}{2}\,$ .
Basel problem : 
The coefficient for $\,z\,$ of the infinite product $\,(1+a_1 z)(1+a_2 z)...\,$ is $\,a_1+a_2+...\,$ . 
Therefore with $z:=-x^2$ and using the series of sine we get $\,\displaystyle\frac{1}{3!}=\frac{1}{(1\cdot \pi)^2}+\frac{1}{(2\cdot \pi)^2}+...\,$ .

Both calculations are based on partial fraction expansion or equivalent on the sine product :
$($ Wallis product $\displaystyle )^2 \,\,= \left(\prod\limits_{k=1}^\infty \left(1-\left(\frac{x}{k}\right)^2\right)^{-1}|_{x=\frac{1}{2}}\right)^2=\left(\frac{\pi x}{\sin(\pi x)}|_{x=\frac{1}{2}}\right)^2=\frac{\pi^2}{4}$
$\displaystyle \frac{\pi^2}{6}=\frac{1-\pi x \cot(\pi x)}{2 x^2}|_{x\to 0}=\sum\limits_{k=0}^\infty \frac{1}{k^2-x^2}|_{x=0}=\zeta(2) $
Both calculations can be put together to an equation chain. The core thing is:
$\displaystyle \frac{2}{3}\prod\limits_{k=1}^\infty \left(1-\left(\frac{x+\frac{1}{2}}{k}\right)^2\right)^{-2} \bigg|_{x=0} =\frac{\pi^2}{6} = -\frac{1}{2}\frac{d^2}{dx^2}\ln \prod\limits_{k=1}^\infty \left(1-\left(\frac{x}{k}\right)^2\right)\bigg|_{x=0} $
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