The inequality is equivalent to
$$
\sin\frac{\sqrt2}{7}<\log\frac{11}{9}.
$$
We use the facts that the taylor series for $\sin x$ and $\log(1+x)$ are alternating and their terms are decreasing in absolute value.
First step: bound $\sin(\sqrt2/7)$ from above.
$$
\sin\frac{\sqrt2}{7}<\frac{\sqrt2}{7}-\frac{1}{3!}\Bigl(\frac{\sqrt2}{7}\Bigr)^3+\frac{1}{5!}\Bigl(\frac{\sqrt2}{7}\Bigr)^5=\sqrt2\,\frac{23847}{168070}<\frac{99}{70}\,\frac{23847}{168070}<0.20067
$$
Second step: bound $\log(11/9)$ from below.
$$
\log\Bigl(1+\frac29\Bigr)>\sum_{n=1}^8\frac{(-1)^{n+1}}{n}\Bigl(\frac29\Bigr)^n=\frac{302337356}{1506635235}>0.20067
$$
I used Mathematica to do the computations, but they can be done by hand in a reasonable time.