Prove that

$\mathrm{tan}x<\frac{4}{\pi}x,\mathrm{\forall}x\in (0;\frac{\pi}{4})$

$\mathrm{tan}x<\frac{4}{\pi}x,\mathrm{\forall}x\in (0;\frac{\pi}{4})$

vasorasy8
2022-07-04
Answered

Prove that

$\mathrm{tan}x<\frac{4}{\pi}x,\mathrm{\forall}x\in (0;\frac{\pi}{4})$

$\mathrm{tan}x<\frac{4}{\pi}x,\mathrm{\forall}x\in (0;\frac{\pi}{4})$

You can still ask an expert for help

Elias Flores

Answered 2022-07-05
Author has **24** answers

Let g be the function defined on the interval $[0,\pi /4]$ as

$g(x)=\{\begin{array}{ll}\frac{\mathrm{tan}x}{x}& ,x\ne 0\\ \\ 1& ,x=0\end{array}$

Then, the derivative g′ of g is given by

${g}^{\prime}(x)=\frac{x{\mathrm{sec}}^{2}x-\mathrm{tan}x}{{x}^{2}}=\frac{x-\frac{1}{2}\mathrm{sin}(2x)}{{x}^{2}{\mathrm{cos}}^{2}x}>0$

for x>0 and g′(0)=0.

Inasmuch as g is increasing for $x\in [0,\pi /4]$ it attains, therefore, its maximum there at $x=\pi /4$

Thus,

$g(x)<g(\pi /4)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{\mathrm{tan}x}{x}<\frac{1}{\pi /4}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{tan}x<\frac{4}{\pi}x$

and we are done!

$g(x)=\{\begin{array}{ll}\frac{\mathrm{tan}x}{x}& ,x\ne 0\\ \\ 1& ,x=0\end{array}$

Then, the derivative g′ of g is given by

${g}^{\prime}(x)=\frac{x{\mathrm{sec}}^{2}x-\mathrm{tan}x}{{x}^{2}}=\frac{x-\frac{1}{2}\mathrm{sin}(2x)}{{x}^{2}{\mathrm{cos}}^{2}x}>0$

for x>0 and g′(0)=0.

Inasmuch as g is increasing for $x\in [0,\pi /4]$ it attains, therefore, its maximum there at $x=\pi /4$

Thus,

$g(x)<g(\pi /4)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\frac{\mathrm{tan}x}{x}<\frac{1}{\pi /4}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{tan}x<\frac{4}{\pi}x$

and we are done!

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