I am asked to prove that $\displaystyle \frac{2}{\pi} < \frac{sin x}{x} < 1$ if $\displaystyle 0 < x < \frac{\pi}{2}$, where the definition of $\displaystyle sin$ is given by $\displaystyle sin(x) = \frac{1}{2}(e^{ix} + e^{-ix})$. I think I have a bit of the problem figured out. I reason that I can solve the whole problem if I know the derivative of $\displaystyle \frac{sin x}{x}$ is always negative, which it is if I am to trust a graphing calculator.

Once that's obtained, I think the rest follows from: $\displaystyle \displaystyle \lim_{x \rightarrow 0^{+}} \frac{\sin x}{x} = \lim_{x \rightarrow 0^{+}} cos(x) = 1$, and $\displaystyle \frac{sin(\frac{\pi}{2}}{\frac{\pi}{2}} = \frac{2}{\pi}$.

So I need to show that $\displaystyle \frac{x cos x - sin x}{x^{2}}$ is negative within these bounds. But I don't see how to do so, or if I need to re-consider how I'm approaching the task.