# Taylor development I don't understand

1. May 3, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! In the middle of an exercise, our teacher suddenly wrote:

$$sin(\frac{x}{y} sin y) = \frac{x}{y} sin y - \frac{1}{2} sin θ (\frac{x}{y} sin y)^2$$

I don't get where does that come from? The closest I've managed to reach is:

$$sin(\frac{x}{y} sin y) = \frac{x}{y} sin y - (\frac{x}{y} sin y)^2 \frac{x sin y}{6y} + O(x^5)$$

What is this θ? It doesn't show up neither before nor after that line :/

Julien.

2. May 3, 2016

### Ray Vickson

When using TeX/LaTeX, please try to remember to use "\sin" instead of "sin", because using "sin" looks ugly and hard to read, like this $sin x$, while using "\sin" produced pleasing and easy-to-read results like this: $\sin x$. Same for cos, tan, csc, sec, cot, (plus their inverses), exp, sinh, cosh, tanh, log, ln, lim, max, min---basically, all the standard functions.

Anyway, he/she is using the standard form of remainder in a Taylor series:
$$f(x) = f(a) + (x-a) f'(a) + \int_a^x f''(t) (x-t) \, dt$$
If $f''$ does not change sign in $[a,x]$ the last term can be expressed as
$$\frac{1}{2!} (x-a)^2 f''(\theta), \; \theta \in [a,x]$$
using the Mean Value Theorem.