Can $\cos(36^\circ)$ be calculated using derivatives?

Question

Can $\cos(36°)$ be calculated using known cosines like $\cos(37°)$ and the derivative formula $$f'(x) \approx \frac{f(x+ Δx) - f(x)}{\Delta x}$$ ?

(We have to let $f(x):=\cos(x)$, so we can let $x=37°$, or in radian form, and let $f(x+\Delta x)$ be $\cos(x+\Delta x)$ and $\Delta x=-1°$ or in radian form, so we can say $\cos(37°-1°)$. I just do not know the process. I understand it is approximate, but I cannot calculate it approximately.

Answer 1

If we know that $\cos{(37^\circ)}\approx 0.8$ and $\sin{(37^\circ)}\approx 0.6$ then we can use the fact that $$\frac{d}{dx} \cos{(x)}=-\sin{(x)}$$ to approximate $\cos{(36^\circ)}$ as $$\cos{(36^\circ)}\approx 0.8+0.6(\frac\pi{180}) = 0.810...$$ Where $1^\circ = \frac\pi{180}$ radians which must be used here.