How does Taylor series work for sine and cosine?

I was just wondering if anyone can help me understand taylor series for sine and cosine. I have no background in calculus but I always found it interesting how the ratios of the arc to the radius was converted to the ratio of sides of a triangle which has straight lines by using Taylor series. Now I totally understand how to calculate Taylor series I just don’t understand why it works. If someone can enlighten me why we keep adding and subtracting radians to different powers divided by factorials why this creates a more and more accurate number. I’m a visual person so a graph or diagram would be helpful. I don’t have any calculus background so going. Through the function in calculus won’t help much. Please also emphasize why we divide by factorials that to me makes no sense to me. Thank you.

• If you have not background in calculus it is difficult to understand why it works (you need to understand the proof and that requires a knowledge of the basic calculus concepts such as derivatives and limits). – gimusi Jun 29 '18 at 14:50
• We divide by factorials because that cancels out the effect of the power rule when we inevitably iterate through differentiation. The goal is to make every derivative of the Taylor series match up with every derivative of the target function. – Andrew Li Jun 29 '18 at 14:58
• Related (duplicate?): "Deriving the power series for cosine, using basic geometry". In particular, see my answer. (There's one element of Calculus needed, to guarantee that certain lengths are what we say they are.) – Blue Jun 29 '18 at 15:07
• Just a word to the physicist in you: don’t be worried about taking the square of an angle, and higher powers, too, because $1$ radian is a dimensionless quantity. – Lubin Jun 29 '18 at 16:35
• It is not possible to make it much simpler than what Allawonder stated in his answer. I gave a sketch here of the intuitive motivation behind the exponential function, namely trying to find a function $\exp$ that satisfies $\exp' = \exp$ and $\exp(0) = 1$. This is motivated by the differential equation for exponential decay. Similarly we can intuitively find the power series for $\cos$ and $\sin$, each of which is a solution $f$ to $f'' = -f$, which is motivated by the differential equation for harmonic motion. But intuition is just one step. – user21820 Nov 5 '18 at 9:01

To really understand why a particular mathematical method works, you need to seriously dig into the proofs -- and best as ever, try to construct your own proof. In this case, you need to understand a very famous result known as Taylor's theorem. Lagrange called it the backbone of all calculus (not his exact words, of course, but something along that line, if not more exaggerated). In summary, if a function has derivatives up to some order $n+1$ about any point of an open interval, then it can be approximated about a point of that interval (in a well-defined sense) by the $n$th order Taylor polynomial with an error that can be estimated well.
The rock bottom issue is not the Taylor series, but the notion of "angle". Note that elementary Euclidean geometry does not tell you what the "measure" $\alpha$ of an angle is, but it has all sorts of theorems about sines, cosines, etc. of such "angles" in store. Such theorems can be proven via Pythagoras' theorem, or area considerations (like the sine theorem), but they don't use the measure of angles per se.
Contrasting this the function $t\mapsto\sin t$ is working with the actual "angle measure". It is not possible to understand the Taylor series of this function without entering the "transcendental" world.