This might take some time

Let $x= \sin 7.5^\circ$ . Then,

$\begin{aligned} 1-2x^2 & = \cos 15^\circ \\ 2\left(1-2x^2\right)^2 - 1 & = 2 \cos^2 15^\circ - 1 \\ 2\left(4x^4-4x^2+1\right) - 1 & = \cos 30^\circ = \frac {\sqrt 3}2 \\ x^4 - x^2 + \frac {2 - \sqrt 3}{16} & = 0 \end{aligned}$

$\begin{aligned} \implies x^2 & = \frac {1 {\color{#3D99F6}\pm} \sqrt{1-4\cdot \frac {2-\sqrt 3}{16}}}2 \\ x & = \sqrt{\frac {2 \ {\color{#D61F06} -} \sqrt{2+\sqrt 3}}4} & \small \color{#D61F06} x = \sin 7.5^\circ < 1 \\ \implies \sin 7.5^\circ & = \sqrt{\frac {2 - \sqrt{\frac {(\sqrt 3+1)^2}2}}4} \\ & = \sqrt{\frac {2\sqrt 2 - \sqrt 3-1}{4\sqrt 2}} \\ & = \boxed{\sqrt {\frac {4-\sqrt 6 - \sqrt 2}8}} \end{aligned}$