Wrong wolfram alpha result for $\sin(\frac{3 \pi}{2})$

Question

I'm trying to compute $\sin(\frac{3 \pi}{2})$ using sin infinite product on wolfram alpha yet it gives me the answer 1 . Shouldnt the answer be -1? here is the result wolfram alpha gives me image

Here is the infinite product formula for $\sin(x)$. Sin(x) infinite product image

Answer 1

The product $$ \frac{3\pi}{2}\prod_{n=1}^\infty\left(1-\frac{(3\pi/2)^2}{n^2\pi^2}\right) $$ has one negative factor $$ \left(1-\frac{(3\pi/2)^2}{1^2\pi^2}\right) $$ and all other factors positive. So that infinite product certainly does not converge to $1$.