C. Inequalities

Inequality 1

$\displaystyle \cos{x} \leq 1-\left({\textstyle\frac{2}{\pi}}{x}\right)^2 \quad \forall {x}\in{\textstyle\left[-\frac{\pi}{2},\frac{\pi}{2}\right]}$

(C.1)

Proof. According to the Weierstraß factorization theorem, the cosine function can be written as [101]

$\displaystyle \cos{x} = \prod_{n=1}^\infty\left(1-\frac{4x^2}{\pi^2\left(2n-1\right)^2}\right).$

(C.2)

Bearing in mind that all factors are in the range $\left[0,1\right]$ for ${x}\in{\textstyle\left[-\frac{\pi}{2},\frac{\pi}{2}\right]}$ , the inequality follows directly by neglecting all factors with $n\geq2$ . The first factor is identical to the right-hand side in (C.1). $\qedsymbol$

Inequality 3

$\displaystyle 1\leq\frac{2\left(\sin\left(\frac{\pi}{2}{x}\right)-{x}\right)}{{x}\left(1-{x}^2\right)} \quad \forall{x}\in\left]0,1\right[$

(C.4)

Proof. The denominator is positive for ${x}\in\left]0,1\right[$ . Thus, the inequality is equivalent to the next statement which is proved in the following for all ${x}\in\left[0,1\right]$ :

	$\displaystyle \frac{{x}\left(3-{x}^2\right)}{2}$	$\displaystyle \leq\sin\left(\frac{\pi}{2}{x}\right)$	(C.5)
$\displaystyle \Leftrightarrow$	$\displaystyle \frac{{x}^2\left(3-{x}^2\right)^2}{4}$	$\displaystyle \leq1-\left(\cos\left(\frac{\pi}{2}{x}\right)\right)^2$	(C.6)
$\displaystyle \Leftrightarrow$	$\displaystyle \left(\cos\left(\frac{\pi}{2}{x}\right)\right)^2$	$\displaystyle \leq\left(1-{x}^2\right)^2\left(1-\frac{{x}^2}{4}\right)$	(C.7)

Using the product representation of the cosine function (C.2):

$\displaystyle \Leftrightarrow$		$\displaystyle \prod_{n=1}^\infty\left(1-\frac{{x}^2}{\left(2n-1\right)^2}\right)^2$	$\displaystyle \leq\left(1-{x}^2\right)^2\left(1-\frac{{x}^2}{4}\right)$	(C.8)
$\displaystyle \Leftrightarrow$		$\displaystyle \prod_{n=2}^\infty\left(1-\frac{{x}^2}{\left(2n-1\right)^2}\right)^2$	$\displaystyle \leq1-\frac{{x}^2}{4}$	(C.9)

Since all factors of this product series are in the range $\left[0,1\right]$ , it is sufficient to show that:

$\displaystyle \Leftarrow$		$\displaystyle \left(1-\frac{{x}^2}{9}\right)^2\cdot\left(1-\frac{{x}^2}{25}\right)^2$	$\displaystyle \leq1-\frac{{x}^2}{4}$	(C.10)
$\displaystyle \Leftrightarrow$		$\displaystyle \left(1-{x}^2\left(\frac{1}{9}+\frac{1}{25}\right)+{x}^4\frac{1}{9\cdot25}\right)^2$	$\displaystyle \leq1-\frac{{x}^2}{4}$	(C.11)

Using ${x}^4\leq{x}^2$ for all ${x}\in\left[0,1\right]$ :

$\displaystyle \Leftarrow$	$\displaystyle \left(1-{x}^2\left(\frac{1}{9}+\frac{1}{25}-\frac{1}{9\cdot25}\right)\right)^2$	$\displaystyle \leq1-\frac{{x}^2}{4}$	(C.12)
$\displaystyle \Leftrightarrow$	$\displaystyle \left(1-{x}^2\frac{11}{75}\right)^2$	$\displaystyle \leq1-\frac{{x}^2}{4}$	(C.13)
$\displaystyle \Leftrightarrow$	$\displaystyle 1-{x}^2\frac{22}{75}+{x}^4 \frac{11^2}{75^2}$	$\displaystyle \leq1-\frac{{x}^2}{4}$	(C.14)

Using again ${x}^4\leq{x}^2$ :

$\displaystyle \Leftarrow$	$\displaystyle 1-{x}^2\left(\frac{22}{75}-\frac{11^2}{75^2}\right)$	$\displaystyle \leq1-\frac{{x}^2}{4}$	(C.15)
$\displaystyle \Leftrightarrow$	$\displaystyle 1-{x}^2\frac{1529}{5625}$	$\displaystyle \leq1-\frac{{x}^2}{4}$	(C.16)
$\displaystyle \Leftarrow$	$\displaystyle \frac{1529}{5625}$	$\displaystyle >\frac{1}{4}$	(C.17)

$\qedsymbol$