Using a Taylor polynomial to calculate the error in using $x$ as an approximation for $\sin{x}$ I'm trying to calculate the error involved in using $x$ to approximate $\sin({1^\circ})$. The Taylor polynomial is centered around $x=0$ (i.e. $x_0=0$), by the way.
Since I have to use $x$ as the approximation for $\sin({x}),$ I'm guessing the Taylor polynomial will have to be of degree one:
$$f(x)=e^x=P_1(x)+R_1(x)$$
since the polynomial $P_1(x)$ does result in $f(x_0)+f'(x_0)(x-x_0)=\sin({0})+\cos({0})x=x$.
When calculating the remainder (i.e. $R_1(x)$), I get that it is given by:
$$\frac{f^{(1+1)}(z)}{(n+1)!}(x-x_0)^2$$ where $z$ is a real number between $x$ and $x_0$. Plugging in the information for this case, then, $$R_1(x)=\frac{f''(z)}{2!}x^2$$ Since $f(z)=\sin({z})$, $f''(z)=-\sin({z})$. The error is then given by $|R_1(x)|=\lvert-\frac{1}{2}x^2 \sin{(z)}\rvert $. 
Trying to find the boundaries for the error specifically when $x=1$ (since the question asks for the error in using $x$ to approximate $\sin(1^\circ)$) gets me that the lower bound is zero (if the estimation were perfectly accurate, which it isn't). I'm having trouble finding the upper boundary, though. The maximum value for |$\sin{(z)}$| is 1 (since the absolute value of the sine function oscillates between 0 and one). Since I'm being asked to calculate the error for when $x=1$, I think $x$ is supposed to be $1$. This would mean that the upper bound would be given by: |$-\frac{1}{2}(1)^2 $|$=\frac{1}{2}$. The given answer, though, is that the error is approximately $8.86 \cdot 10^{-7}$.
What did I do wrong? It always struck me that $x$ would be an unsuitable approximation for $\sin{(1^\circ)}$ since, at $x=1$, $\sin({1^\circ})$ is very small, certainly less than one (the maximum value for the sine function). I don't understand how the error can be so relatively small.
 A: The error is that you are plugging in $x=1$, instead of $x=1^\circ$. A degree is $\frac{\pi}{180}$ radians, so you should set $x=\frac{\pi}{180}$ instead.
(The right way to think about this is that the sine function is just defined by $\sin x=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$.  When $x$ is an angle measure in radians, then $\sin x$ defined in this way happens to also coincide with the geometric "sine" function you learn about in trigonometry.  But if you measure $x$ in degrees it does not.)
Also, you can get a much better upper bound for the $\sin(z)$ part.  Since $z$ is between $0$ and $1^\circ$ and sine is increasing on this interval, the most that $\sin(z)$ could be is $\sin 1^\circ=\sin\frac{\pi}{180}$.  Of course, you are trying to estimate $\sin\frac{\pi}{180}$, so you don't yet have an exact value for this that you can plug in.  But you can use this to iteratively get a better bound for the error: once you have one bound for the error by saying $|\sin(z)|\leq 1$, you can then plug in the bound on $\sin \frac{\pi}{180}$ you got as a new bound on $|\sin z|$ to get a better estimate of the error.
