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