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I want to prove that $\frac{\sin t}{t}$ is concave in $[0,\frac{\pi}{2}]$,but when I differentiate twice, I get

$$-\frac{\sin t}{t}-2\frac{\cos t}{t^2}+2\frac{sint}{t^3}$$

then simplify it, I want to prove

$-\tan (t)t^2-2t+2tan(t)<0$,but I can't continue,can anyone give a help?

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up vote 2 down vote accepted

For $\sqrt{2}\le t \le \pi/2$ this is clearly true. For $0 < t < \sqrt{t}$, note that $\sin(t) < t$ and $\cos(t) > 1 - t^2/2$.

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