# Using mean value theorem to show that $\sin x\lt x$

What I dont understand is this, we are stating that there exists at least one c on the open interval
$(0,2 \pi )$, such that the slope of the chord line $\frac{\sin (x)-\sin (0)}{x+0}$ is equal to the slope of the the tangent line $\frac{d \sin (x)}{\text{dx}}$, and we are denoting this point with the name c. However, how does two or more parallel lines prove that $\sin (x)<x$ for all x>0. And why does the slope at point c, -> have to be less than one?

• The solution is just equating the slopes. And the slope is less than $1$ since $\cos c$ is less than $1$ between $0$ and $2\pi$. – Nishant Aug 4 '14 at 4:19
• How do I know that cos[c] is less than 1 between 0 and $2 \pi$ – ALEXANDER Aug 4 '14 at 4:22
• $\cos^2 x + \sin^2 x = 1$ so $\cos x \leq 1.$ Are you worried about strict inequality? – RRL Aug 4 '14 at 4:27
• Yes I just do not get it! – ALEXANDER Aug 4 '14 at 4:29

For each $x \in (0,2\pi),$ the mean value theorem ensures that there exists $c_x$ with $0 < c_x < x$ such that

$$\frac{\sin x}{x} = \cos c_x < 1.$$

Mean Value Theorem: If $f:[a,b] \rightarrow \mathbf{R}$ is differentiable in $(a,b)$, there exists $c$ with $a < c < b$ such that $f(b)-f(a) = f'(c)(b-a).$

With $f(x) = \sin x$ we have $\sin x = \sin x - \sin 0 = f'(c_x)(x-0)= (\cos c_x)x$ where $0 < c_x < x$.

• @ALEXANDER: Which part do you not get? Mean value theorem used above or $\cos x \leq 1$? – RRL Aug 4 '14 at 4:34
• I do get the $\cos (x)\leq 1$, but I do not get the Mean Value Theorem – ALEXANDER Aug 4 '14 at 4:35
• Its that the average change over time is equal to the instant change over time at least once in the interval. And since this slope is less than one on this interval how does that prove that Sin[x] is less than x. There is something obvious that I am missing! – ALEXANDER Aug 4 '14 at 4:41
• I quoted the MVT and applied it to this case above. If its not clear look up Mean Value Theorem on Wikipedia etc for a proof. It shows for each $x$ that $\sin x / x$ equals $\cos$ at some point between $0$ and $x$. You can show $\cos$ is strictly less than 1 on $(0,2\pi)$. – RRL Aug 4 '14 at 4:47
• First line "For each ...". You are confusing the fixed $c$ that matches secant and tangent for one fixed interval say $x=0$ to $x=1$ with the $c_x$ that works on the interval $[0,x]$. Say $x = 0.9$, then $\sin0.9/0.9 = \cos(0.54857...) = 0.87036...$. So $\sin0.9 = (0.87036...)(0.9) < 0.9$. Say $x = 0.5$, then $\sin0.5/0.5 = \cos(0.28786...) = 0.95885...$. So $\sin0.5 = (0.95885...)(0.5) < 0.5$. – RRL Aug 4 '14 at 10:51