Prove that for $x>0$, $\sin x1-\frac{x^2}{2}$ Please, how do I solve this problem? Do I have to use differentiablity which we just did today or what? Here is the question. I don't know how to go about it.
Prove that for $x>0$, $\sin x<x$ and $\cos x>1-\frac{x^2}{2}.$
Please, can anyone help give a detailed explanation?
 A: Consider the small right triangle with legs $\sin x, 1-\cos x$

The length of the vertical $(\sin x)$ is less than the hypotenuse (lets call it $h$), which is less than the length of the arc $(x).$
When $x> 0$
$\sin x < h < x$
Use the Pythagorean theorem to find $h$
$\sin^2 x + (1-\cos^2x) = h^2 < x^2\\
2-2\cos x < x^2\\
\cos x > 1-\frac {x^2}2$
Another way to do it....
$\sin x = $$\sum_\limits{n=0}^{\infty} (-1)^n \frac {x^{2n+1}}{(2n+1)!}\\
 x + \sum_\limits{n=1}^{\infty} (-1)^n \frac {x^{2n+1}}{(2n+1)!}$
and
$\cos x = $$\sum_\limits{n=0}^{\infty} (-1)^n \frac {x^{2n}}{(2n)!}\\
 1 - \frac {x^2}{2} + \sum_\limits{n=2}^{\infty} (-1)^n \frac {x^{2n+1}}{(2n+1)!}$
These series are are alternating series.  Which means that the partial sums alternate from below to above sin $x$ and cosine $x$, respectively, with each successive term.
A: HINT: Use $f(0)=g(0)$ and $f'(x)\ge g'(x)$ to deduce that $f(x)\ge g(x)$ for $x\ge 0$. You can use either the mean value theorem or integrals. You can get inequality (rather than allowing equality) in your case with a smidgeon more work.
A: Hint:
$D(\sin y)=\cos y \leq 1$ , $\sin 0=0$ and use mean value theorem to obtain $\sin y <y$ for $y>0$
A: For the first part, note that is x is a point on the unit circle then  $x$ is  the length of the arc from the positive direction of the x-axis to the point while  $\sin x $ is  the length of the perpendicular segment dropped from the point x on the circle to the x-axis. The arc has a larger length than the perpendicular. 
For the second part use binomial theorem to show $$ \sqrt {1-t^2}=1-\frac {t^2}{2} + \frac {t^4}{8}..>1-\frac {t^2}{2}$$ with $t=\sin x$ and use $$ 1-\frac {\sin ^2x}{2}> 1-\frac {x ^2}{2}$$
