A Trigonometric Inequality from the RMM


A Trigonometric Inequality from the RMM


Due to the addition formula for sine, $\sin (x+y)=\sin x\cos y+\cos x\sin y,\;$ to prove the required inequality suffice it to establish that, for $\displaystyle z\in\left(0,\frac{\pi}{2}\right),\;$ $\displaystyle\cos z\lt\left(\frac{\sin z}{z}\right)^3.$

This is the same as $f(z)=\sin^2z\tan z - z^3 \gt 0.\;$ Note that $f(0)=0.$ We shall differentiate repeatedly.

$\displaystyle\begin{align} f'(z) &= \sin^2z\,\sec^2z+\tan z(2\sin z\cos z)-3z^2\\ &=\tan^2 z+2\sin^2 z-3z^2. \end{align}$

Introduce $g(z)=f'(z)\;$ and note that $g(0)=0.$

$\displaystyle g'(z)= 2\tan z\,\sec^2z+4\sin z\cos z-6z.$

Introduce $h(z)=\frac{1}{2}g'(z)\;$ and note that $h(0)=0.$

$\displaystyle\begin{align} h'(z) &= \left(\sec^2z\right)^2+(\tan z)(2\sec z)(\sec z\,\tan z)+2(\cos^2z-\sin^2z)-3\\ &=\left(1+\tan^2z\right)^2+2\tan^2z(1+\tan^2z)+2(2\cos^2z-1)-3\\ &=(1+t^2)^2+2t(1+t^2)+\frac{4}{1+t}-5\\ &=\frac{3t^3+7t^2}{1+t}, \end{align}$

$t=\tan z\;\gt 0$ and so $h'(z)\gt 0.\;$ Hence, $h(z)\gt h(0)=0,\;$ so that $g'(z)\gt 0,\;$ and $g(z)\gt g(0)=0,\;$ meaning $f'(z)\gt 0\;$ and $f(z)\gt f(0)=0.$

This completes the proof.


Dan Sitaru has kindly posted the above problem from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page.

The proof is by Soumava Chakraborty.


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