I have found an interesting family of infinite products. The most interesting one of them being:
$\sqrt{2}-1=\dfrac{1\cdot7\cdot9\cdot15\cdot17\cdot23\cdots}{3\cdot5\cdot11\cdot13\cdot19\cdot21\cdots}$
The numerators follow the pattern $+6,+2,+6,+2,\cdots$ and the denominators follow $+2,+6,+2,+6,\cdots$
The family of products was derived from assuming: $\sin(x)-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\cdot\left(1-\dfrac{4x}{\pi}\right)\left(1-\dfrac{4x}{3\pi}\right)\left(1+\dfrac{4x}{5\pi}\right)\left(1+\dfrac{4x}{7\pi}\right)
\cdots$
And subsequently replacing $x$ as $\dfrac{\pi}{2}$
The keyword here being "assumed", so I don't know if this product can be proven. The values do seem to be equal after a few iterations though.
However, I would greatly appreciate any proofs or alternate derivations.
Best Answer
You wrote
That assumption is indeed correct.
Proof: Using the “sum to product” identity for the sine we have $$ f(x) = \sin(x) - \frac{1}{\sqrt 2} = \sin(x) - \sin\left(\frac \pi 4\right) = -2 \sin\left(\frac x2 - \frac \pi 8\right) \sin\left(\frac x2 - \frac {3\pi}8 \right) \, . $$ Using the infinite product for the sine (see, e.g., here), the first factor can be written as $$ \sin\left(\frac x2 - \frac \pi 8\right) = \left(\frac x2 - \frac \pi 8\right)\prod_{n=1}^\infty \left(1-\frac{x}{2\pi n} + \frac{1}{8n}\right) \left(1+\frac{x}{2\pi n} - \frac{1}{8n}\right) \\ = (-\frac \pi 8) \left(1 - \frac {4x}{\pi} \right)\prod_{n=1}^\infty \left(1 + \frac{1}{8n}\right)\left(1-\frac{4x}{(8n+1)\pi} \right) \left(1- \frac{1}{8n}\right)\left(1+\frac{4x}{(8n-1)\pi}\right) \\ = C_1 \left(1 - \frac {4x}{\pi} \right)\prod_{n=1}^\infty \left(1+\frac{4x}{(8n-1)\pi}\right)\left(1-\frac{4x}{(8n+1)\pi} \right) $$ with the constant $$ C_1 = (-\frac \pi 8)\prod_{n=1}^\infty \left(1- \left(\frac{1}{8n}\right)^2\right) \, . $$ In the same way we get for the second factor $$ \sin\left(\frac x2 - \frac {3\pi}8 \right) = C_2 \left(1 - \frac {4x}{3\pi} \right)\prod_{n=1}^\infty \left(1+\frac{4x}{(8n-3)\pi} \right) \left(1-\frac{4x}{(8n+3)\pi}\right) $$ with some constant $C_2$.
Combining these results we have $$ f(x) = C \left(1 - \frac {4x}{\pi} \right)\left(1 - \frac {4x}{3\pi} \right) \\ \times \prod_{n=1}^\infty\left(1+\frac{4x}{(8n-3)\pi} \right) \left(1+\frac{4x}{(8n-1)\pi} \right)\left(1-\frac{4x}{(8n+1)\pi} \right)\left(1-\frac{4x}{(8n+3)\pi} \right) $$ with some constant $C$.
Setting $x=0$ shows that $C= -1/\sqrt 2$, and that concludes the proof.