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".
 A: 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, 
$$\color{blue}{1-\frac{x^2}{2}\le \cos x\le 1-\frac{x^2}{2}+\frac{x^4}{24} }$$
$\vdots$
$$\color{blue}{x-\frac{x^3}{6} \le\sin x\le x-\frac{x^3}{6} +\frac{x^5}{5!}}$$
A: 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 $\sin x \ge x - \frac{x^3}{6}$
A: 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
A: A geometric proof is as follows.
Outline:


*

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

*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 $\sin x\geq x-\frac{1}{6}x^3.$
