Can not derive $\cos (x + y)$

Another approach, since you already proved the sum identity for sines: $$\cos(a+b)=\sin(90^\circ- (a+b))\\ =\sin((90^\circ- a)+(-b))\\ =\sin(90^\circ- a)\cos(-b)+\cos(90^\circ- a)\sin(-b)\\ =\cos(a)\cos(b)-\sin(a)\sin(b) .$$

You nearly got it. You have to use that $DE=1·\sin(1°)$ from the ODE triangle, and also $DE=DF·\cos(10°)$ inside the DEF triangle. Thus $$ EF=DF·\sin(10°)=\frac{\sin(1°)\sin(10°)}{\cos(10°)} $$

$\overline{OE} = \cos(1^\circ)$.

Therefore
$\overline{OM} = \cos(10^\circ)\cos(1^\circ)$.

$\overline{ED} = \sin(1^\circ)$.

Also $\angle DEG = 10^\circ.$

Therefore,
$\overline{MK} = \overline{DG} = \sin(10^\circ)\sin(1^\circ)$.

Finally,
$\cos(11^\circ) = \overline{OK} = \overline{OM} - \overline{MK}.$