# Estimate from below of the sine (and from above of cosine)

I'm trying to do the following exercise with no success. I'm asked to prove that

$$\sin(x) \ge x-\frac{x^3}{2}\,, \qquad \forall x\in [0,1]$$

By using Taylor's expansion, it's basically immediate that one has the better estimate

$$\sin(x) \ge x-\frac{x^3}{6}\,, \qquad \forall x\in [0,1]$$

as the tail converges absolutely, and one can check that the difference of consecutive terms is positive.

I suppose then, there is a more elementary way to get the first one. Question is: how?

Relatedly, the same exercise asks me to prove that

$$\cos(x) \le \frac{1}{\sqrt{1+x^2}}\,,\qquad \forall x\in [0,1]$$

which again I can prove by using differentiation techniques. But these haven't been explained at that point of the text, so I wonder how to do it "elementarly".

I showed by comparison of areas that for first quadrant angles $$\sin\theta\cos\theta\le\theta\le\tan\theta$$ If one multiplies the left of these inequalities by $2$ it becomes $\sin2\theta<2\theta$ so we arrive at $$\sin\theta\le\theta\le\tan\theta$$ Rearrange the right of these inequalities to $$\frac{\sin\theta}{\theta}\ge\cos\theta$$ or $$1-\frac{\sin\theta}{\theta}\le1-\cos\theta=2\sin^2\frac{\theta}2\le2\left(\frac{\theta}2\right)^2=\frac{\theta^2}2$$ Where we have used the left of the above inequalities above. This rearranges to $$\sin\theta\ge\theta-\frac{\theta^3}2$$ for first quadrant angles.

EDIT: I missed the second part of your question! The left of the inequalities can be squared into $$\tan^2\theta\ge\theta^2$$ or $$\sec^2\theta=\tan^2\theta+1\ge\theta^2+1$$ Taking reciprocals and square roots, $$\cos\theta\le\frac1{\sqrt{\theta^2+1}}$$ again for first quadrant angles

You can use Mean Value Theorem on the function $f(x) = \sin(x) - x + \frac{x^3}{2}$. Then we have that $\exists c \in (0,x)$ s.t.

$$\frac{f(x) - f(0)}{x-0} = f'(c) \iff \sin(x) - x + \frac{x^3}{2} = x\left(\cos(c) - 1 + \frac{3c^2}{2}\right)$$

Now using Mean Value Theorem on $g(x) = \cos(x) - 1 + \frac{3x^2}{2}$ we have that $\exists c_1 \in (0,c)$ s.t.

$$\frac{f(c) - f(0)}{c-0} = f'(c_1) \iff \cos(c) - 1 + \frac{3c^2}{2} = c(-\sin(c_1) + 3c_1) \ge 0$$

Combining these we have that $\sin x \ge x - \frac{x^3}{2}; \forall x \in [0,1]$. Moreover you can prove the stronger inequality you just mentioned, namely $\sin x \ge x - \frac{x^3}{6}$

We know that $$\sin t = \int_0^t\cos s ds \le t$$ for $0\lt t\lt z\lt x\lt 1$. Integrating over $\color{blue}{(0,z)}$ we get

$$1-\cos z=\int_0^z\sin tdt \le \int_0^ztdt= \frac{z^2}{2}$$ that is for all $0<z<x<1$ we have, $$\color{blue}{1-\frac{z^2}{2}\le \cos z\le 1}$$ integrating again over $\color{blue}{(0,x)}$ we get

$$\color{red}{x-\frac{x^3}{6} = \int_0^x 1-\frac{z^2}{2} dz\le \int_0^x\cos z dz=\sin x}$$ that is $$\color{blue}{x-\frac{x^3}{6} \le\sin x\le x}$$

continuing with this process you get, $$\color{blue}{1-\frac{x^2}{2}\le \cos x\le 1-\frac{x^2}{2}+\frac{x^4}{24} }$$ $\vdots$

$$\color{blue}{x-\frac{x^3}{6} \le\sin x\le x-\frac{x^3}{6} +\frac{x^5}{5!}}$$

• Thanks for taking your time, but integrating is in no way `more elementary' than using differentation techniques.. – Paolo Intuito Dec 7 '17 at 16:09
• @G.S. If you say so. for me this is more easiest way ever:) – Guy Fsone Dec 7 '17 at 16:25

A geometric proof is as follows.

Outline:

1. Show $\cos x > 1-\frac{1}{2}x^2.$
2. Show that $\tan x> x.$

From there, you quickly see that $\sin x>x\cos x>x-x^3/2.$

We have that $\sqrt{(1-\cos x)^2+\sin^2 x}$ is the length of segment from $(1,0)$ to $(\cos x,\sin x)$, which is $\leq x$, since the arc along the circle between these two points is length $x$ and the shortest distance between two points is the line.

But $(1-\cos x)^2+\sin^2 x=2-2\cos x$.

So you have $2-2\cos x < x^2$, or $\cos x > 1-\frac12x^2$.

The area of the triangle $(0,0),(1,0),(1,\tan x)$ is $\frac{1}{2}\tan x$ and this triangle contains a region of the unit circle of area $\frac{1}{2}x$. So you get that $\tan x>x.$

This gives the result you want.

The second part uses that $\tan^2 x + 1=\frac{1}{\cos^2 x}$ so $$\cos x = \frac{1}{\sqrt{1+\tan^2 x}}<\frac{1}{\sqrt{1+x^2}}$$ since $x<\tan x$.

Extending to show $\sin x > x-x^3/6$:

Now, if, for all $x\in(0,\pi/2)$, $x>\sin x>x-ax^3$ for $x\in (0,\pi/2)$ then use:

$$\sin x = 2\sin \frac{x}{2}\cos\frac{x}{2}>2\left(\frac{1}{2}x-\frac{a}{8}x^3\right)\left(1-\frac{1}{8}x^2\right)>x-\frac{2a+1}{8}x^3$$

This requires both $1-\frac{a}{4}x^2$ and $1-\frac{1}{8}x^2$ to be positive. Since all values of $a$ in question will be $\leq \frac12$ we want $x<\sqrt{8}$, which is clearly true for $x\in(0,\pi/2).$

If we define $a_0=\frac{1}{2}$ and $a_{n+1}=\frac{2a_n+1}{8}$ then you have that $a_n$ is decreasing and the limit is $\frac{1}{6}.$ Since the $0<a_n\leq \frac{1}{2}$, we get, inductively, for $x\in(0,1)$ that:

$$\sin x > x-a_nx^3$$

In the limit, this means that $\sin x\geq x-\frac{1}{6}x^3.$