Approximating trig functions through Taylor / Maclaurin I have a test on Tuesday and I'm trying to review a section  I wasn't here for during class and I'm really confused. 
I know how to find taylor and maclaurin polynomials but there is a question that asks me to approximate $\sin(4^\circ)$ to five decimal places.
I know I need to turn it into radians which would be $\frac{\pi}{45}$ and I know the polynomial would be: $$\frac{\pi}{45} - \frac{(\frac{\pi}{45})^3}{3!}$$ and then alternates with odd exponents and odd factorials.
What would the next step be though? I know it has something to do with the remainder theorem but I'm kinda lost. Thanks a lot.
 A: Well firstly, $\sin(x) \approx x$ when $x$ is very small. You can see this by taylor expansion of $\sin(x)$ as you did in the question, and noticing that $x^n, n > 1$ goes to $0$ very fast. $4$ degrees is quite close to zero, so you could conclude that 
$$\sin(4^\circ) = \sin(\frac{\pi}{45}) \approx \frac{\pi}{45} \approx 0.069813$$
And we are done if we didn't need a specific number of decimal places. Using a calculator, you would find $\sin(\frac{\pi}{45}) \approx 0.069756$, so our guess was pretty good.
To get the correct precision, we will use the remainder formula, which is: When using an $n^{\text{th}}$ degree Taylor polynomial $P_n(x)$ centered at $a$ the error in approximating $f(x)$ is bounded by $$R_n = \left|\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\right|, \ c \in (a,x) $$
Let $x = \frac{\pi}{45}$ and center at $a=0$ we want $n$ such that $R_n < 0.00001$. For $n=1$ 
$$R_1 \leq \frac{(\frac{\pi}{45})^2}{2!} \approx 0.002, \text{ Since } \frac{d^2}{{dx}^2}\sin(x) = -\sin(x) \in [-1,1] $$
So $n=1$ is not good enough, as we already found when we guessed using $\sin(x) \approx x$. The next $n$ to try would be $n=3$, since $P_1(x) = P_2(x)$. You should find $R_3 \leq 0.00001$ so that 
$$\sin(4^{\circ}) = P_3(\frac{\pi}{45}) = \frac{\pi}{45} - \frac{(\frac{\pi}{45}^3)}{3!} \approx 0.069756$$
gives you an accurate enough approximation. 
