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Let

$$ 1< \alpha_1<\alpha_2<......<\alpha_n< \pi/2$$ then prove that $$ tan(\alpha_1)< \frac{sin(\alpha_1)+sin(\alpha_2)+.....+sin(\alpha_n)}{cos(\alpha_1)+cos(\alpha_2)+......+cos(\alpha_n)} < tan(\alpha_n)$$

I tried to solve the problem but I am not actually able to get a well established answer or you can say I am not even able to advance ahead to solve this one. It tried to use AM-GM inequality, but its didn't work as I couldn't establish any direct relation.

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  • $\begingroup$ It's easy, $\sin(x)$ function is ascending on the given interval, while $\cos(x)$ is descending. $\endgroup$ – rtybase Jun 4 '16 at 9:33
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Because $\sin(x)$ is ascending function on the given interval: $$n\cdot \sin(\alpha_1) < \sin(\alpha_1) + ... + \sin(\alpha_n) < n\cdot \sin(\alpha_n)$$

Because $\cos(x)$ is descending function on the given interval: $$n\cdot \cos(\alpha_1) > \cos(\alpha_1) + ... + \cos(\alpha_n) > n\cdot \cos(\alpha_n)$$

Or: $$\frac{1}{n\cdot \cos(\alpha_1)} < \frac{1}{\cos(\alpha_1) + ... + \cos(\alpha_n)} < \frac{1}{n\cdot \cos(\alpha_n)}$$

All positive values. That's pretty much it.

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