calculating $\sin(32^\circ)$ using the Taylor approximation

Question

The question is:

Compute $\sin(32^\circ)$ (32 degrees) with an error of no more than 0.0001 using the Taylor approximation.

I need to do this without a calculator, and I get stuck with very high numbers. What I tried to do:

Looking for the remainder, by taylor remainder theorem

$$|R(32)| = \left|\frac{f^{(N+1)}(c)\cdot 32^{N+1}}{(N+1)!}\right| < \frac{32^{N+1}}{(N+1)!}$$

Now I'm looking for $N$ so:

$$\frac{32^{N+1}}{(N+1)!} < 0.001$$

But It doesn't make any sense. I'm dealing with very large numbers.

Answer 1

$$2^o=2 \times \frac{2\pi}{360}=\frac{\pi}{90} $$ $$f(x+h)=x+f'(x)\times h+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...\\sin(\frac{\pi}{6}+\frac{\pi}{90})=sin(\frac{\pi}{6})+cos(\frac{\pi}{6})\frac{\pi}{90}-sin(\frac{\pi}{6})(\frac{\pi}{90})^2-cos(\frac{\pi}{6})(\frac{\pi}{90})^3+\cdots $$you need to find $R(2^o)=R(\frac{\pi}{90})$ so $$|R(2^o)|=|R(\frac{\pi}{90})| = \left|\frac{f^{(N+1)}(c)\cdot (\frac{\pi}{90})^{N+1}}{(N+1)!}\right| < \frac{(\frac{\pi}{90})^{N+1}}{(N+1)!}<0.001\to\\ \frac{(0.035)^{N+1}}{(N+1)!}<10^{-3}\\$$now put $N=1,2,3,...$ to find first suit number

Answer 2

Hint: $\sin30°$ and $\cos 30°$ are well-known, you want to compute $\sin(30°+2°)$