The maclaurin series of $\sin(x)$ is $x- x^3/3! + x^5/5! - \cdots + (-1)^n x^{2n+1}/(2n+1)!$.
My teacher wants me to use Taylor's inequality theorem on page 607 to solve this problem.
I know that the alternating series test can apply, but I'm required to use Taylor's theorem for this problem. I think that $M$ is equal to $1$, $a$ is equal to $0$, and so I have $10^{-7}\le (1/10)^{n+1}/(n+1)!$ but I don't know how to solve for n and I'm not sure if that setup is correct.
Note the very useful fact that the Taylor polynomial of $\sin x$ up to and including the term in $x^{2n+1}$ is exactly the same as the Taylor polynomial up to and including the term in $x^{2n+2}$.
The $(2n+2)$-th derivative of $\sin t$ is one of $\cos t$ or $-\cos t$, so it has absolute value $\le 1$. We can therefore take $M=1$.
The absolute value of the error when we cut off at the term in $x^{2n+1}$ is therefore less than $$\frac{1}{(2n+3)!} |x|^{2n+3}.$$
With $x=\frac{1}{10}$, the error quickly gets below $10^{-7}$. Already it is a bit below at $n=1$.
So the approximation $\sin x\approx x-\frac{x^3}{3!}$ is good enough when $x=\frac{1}{10}$.