Prove $\sin(x)\tan(x) > x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$
So I did the following: Let $f(x) = \sin(x)\tan(x) - x^2$.
Then of course $f(0) = 0$ and I want to show that for $x \in ( \,0, \frac{\pi}{2}) \,$ $f'(x) > 0$ but I don't really know where to go from here as I can't get anything reasonable done with the derivative.
I was thinking of approximating it using the following inequality $\sin(x) < x < \tan(x)$.
Is that a good direction?
$\endgroup$25 Answers
$\begingroup$Clearly we can see that $$ \tan(x) ,\cos(x)>0 \, \forall x \in (0,\frac{ \pi}{2}) \, . $$ Now as you said, the derivative is $$ f'(x)=\cos(x)\tan(x)+ \frac{\sin(x)}{\cos^2(x)} -2x \\=\cos(x)\tan(x)+\tan(x)\sec(x)-2x \\ =\tan(x)\left(\cos(x)+\frac{1}{\cos(x)}\right)-2x \geq 0 $$ since $$ \tan(x)>x ,x+\frac{1}{x} \geq 2 \, \forall x>0 \, . $$
$\endgroup$4$\begingroup$Using Geometric mean $\geq $ Harmonic mean
$$\sqrt{\sin x\cdot \tan x}\geq \frac{2}{\frac{1}{\sin x}+\frac{1}{\tan x}}=2\tan \frac{x}{2}>x$$
becsuse $\displaystyle \tan x>x\;\forall \;x\in(0,90^\circ)$
So $$\sin x\cdot \tan x>x^2\;\forall\; x\in (0,90^\circ)$$
$\endgroup$$\begingroup$Let $f(x)=\sin(x)\tan(x)-x^2=\frac{1}{\cos(x)}-\cos(x)-x^2$. Then $f(0)=0$, $f'(0)=0$ and for $x\in (0,\pi/2)$, $$f''(x)=\underbrace{\frac{2\sin^2(x)}{\cos^3(x)}}_{>0}+\underbrace{\frac{1}{\cos(x)}+\cos(x)}_{> 2}-2>0.$$ Then the desired inequality follows easily: $f$ is strictly convex over $(0,\pi/2)$ and therefore its graph lies above the tangent line at $x=0$, i.e. the line $y=0$.
$\endgroup$0$\begingroup$If you can use Taylor series, then you may proceed as follows:
- $\sin(x)\tan(x) > x^2 \Leftrightarrow \boxed{\left(\frac{\sin x}{x}\right)^2 > \cos x}$ on $(0,\frac{\pi}{2})$
Using Taylor you get on $(0,\frac{\pi}{2})$:
- $\frac{\sin x}{x} > 1-\frac{x^2}{6} \Rightarrow \boxed{\left(\frac{\sin x}{x}\right)^2 > 1-\frac{x^2}{3} + \frac{x^4}{36}}$
- $1-\frac{x^2}{2} + \frac{x^4}{24} > \cos x$
Remains to show $1-\frac{x^2}{3} + \frac{x^4}{36} > 1-\frac{x^2}{2} + \frac{x^4}{24}$ on $(0,\frac{\pi}{2})$:
\begin{eqnarray*} & 1-\frac{x^2}{3} + \frac{x^4}{36} > 1-\frac{x^2}{2} + \frac{x^4}{24} &\\ & \stackrel{x > 0}{\Leftrightarrow} & \\ & |x| < 2\sqrt{3} & \end{eqnarray*} Since $\frac{\pi}{2} < 2\sqrt{3}$, we are done.
$\endgroup$$\begingroup$A possible approach: Use Taylor's theorem to show that $$ \sin x > x - \frac{x^3}{6} \\ \tan x > x + \frac{x^3}{3} $$ for $0 < x < \frac\pi 2$. Then $$ \sin x \tan x > x^2 + \frac{x^4}{6} - \frac{x^6}{18} = x^2 + \frac{x^4(3 - x^2)}{18} > x^2 $$ in that interval.
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