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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 $\sin(x) \leq x$?

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Marcos TV
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    "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 $\sin(x)\le x$"

    What you are trying to prove would not prove that $\sin x\le x$ for all $x\in[0,1]$.

     –  May 23 '17 at 06:20
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    You could use the Mean Value Theorem. – Angina Seng May 23 '17 at 06:20
  • Differentiate, and prove that $\cos(x)\leq1$. – Arthur May 23 '17 at 06:21
  • For this question to make any sense, you need to tell how you define the sine and what theory you are allowed to use. –  May 23 '17 at 06:59

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Use differentiation. You know that at $x=0$, $\sin(0) = 0$. But since $\displaystyle \frac{d}{dx} \sin(x) = \cos(x)$, and $\displaystyle \frac{dx}{dx}=1$, you can use the fact that $|\cos(x)| \le 1$ for all $x$. Can you draw the conclusion, then?

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

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