$\sin(x) \leq x$ on the interval $[0,1]$ I need to prove that:  $\sin(x) \leq x$ on the interval $[0,1]$ using Calculus. First I calculated the area between the graph of the function $x$ and $\sin(x)$ on the interval $[0,1]$, if it is positive then $\sin(x) \leq x$:
$$\int_0^1 x - \sin(x)  dx = \frac{x^2}{2} \vert_0^1 +\cos(x) \vert_0^1=\frac{1}{2} - 0 + \cos(1) -1=\cos(1)-\frac{1}{2}$$ 
With the help of the calculator I know that $\cos(1) > \frac{1}{2}$, but I don't know how to prove it without help of the calculator. Any idea of how to prove that $\cos(1) > \frac{1}{2}$ or another way to prove $\sin(x) \leq x$?

Fix $x \in [0,1]$,and write $\sin x = \displaystyle \int_{0}^x \cos t dt, x = \displaystyle \int_{0}^x 1dt\implies x - \sin x = \displaystyle \int_{0}^x (1-\cos t)dt\ge \displaystyle \int_{0}^x 0dt  = 0\implies x \ge \sin x$
