Approximate $\cos(62^{\circ})$.
I think I'm missing something in my calculation, since I don't get an answer I expect.
Since $\cos(60^{\circ}) = \cos(\frac{\pi}{3})$, I'm going to use that as my starting point. The approximate change is
$$\delta y = f'(x) \delta x$$
and $\delta x$ is $2^{\circ} = \frac{\pi}{90}$. Then
$$f'(x) = -\sin(\frac{\pi}{3})$$
so
$$\delta y = -\sin(\frac{\pi}{3}) \frac{\pi}{90} = -\frac{\sqrt3}{2}\frac{\pi}{90} = -\frac{45 \sqrt3}{180} = -\frac{\pi \sqrt3}{4}$$
So then $y + \delta y = \frac{\sqrt3}{2} - \frac{\pi \sqrt3}{4}$, which is not what I'm expecting since
$$\cos(\frac{\pi}{3} + \frac{\pi}{90}) > \cos(\frac{\pi}{3})$$
Where am I going wrong here?
You've got both $y$ and $\delta y$ wrong. $y=\cos(60^\circ)=\frac{1}{2}$, and $\delta y\approx\frac{\pi \sqrt{3}}{180}.$ Not sure where your magic factor of $45$ comes from. So you get:
$$\cos(62^\circ)=y+\delta y\approx\frac{1}{2}-\frac{\pi\sqrt{3}}{180}.$$