Question

prove that cosx>=1-(x^2/2)for all x in the interval[0,2π]

Answer 1

$Consider \;the \;following\;function\;: \\ f(x)= \cos(x) -1 +\dfrac{x^2}{2} \\ Then\;the\;assertion\;that\; \cos(x) \geq 1 -\dfrac{x^2}{2}\; is \;identical \;to\; showing \;that\;: \\ \;\;\;\;\;\;\;\;\;\;\cos(x) -1 +\dfrac{x^2}{2} \geq0 \iff f(x)\geq0$

$When \; x = 0,\; we\; have\; f ( 0 ) = 0.$

$So\;to\;prove\;that\;f(x) \geq 0 \;we\; attempt \;to \;verify\; that\; the\; function\;f(x) \\is \;strictly\;increasing.To \;do \;this, \;take \;the \;derivative:$

$f'(x)=-\sin(x)+x$

$If \;we\; show f ' ( x ) \;is\; strictly\; positive\; for\; x>0, then \;we\; can\; apply\; the \;mean \;value\; \\ theorem \;to\; show\; f ( x )\; must\; be \;strictly\; increasing. \;Differentiating \;again\; we\; get:$

$f''(x)=-\cos(x)+1$

$Clearly , \;as \;\cos(x)\leq 1 \implies 1-\cos(x) \geq0 \;and\;so\;we\;can\;use\;this \\ (and\;mean\;value\;theorem\;again) \;to\;prove\;f'(x)>0\; \forall \;x\isin[0,2\pi],\;hence\;the\;result \\ follows.$