# If x_1,x_2,x_3,......,x_n are n real numbers satisfying 0<x_1<x_2<...<x_n<pi/2, how do you prove that tanx_1 < (sinx_1+sinx_2+...+sinx_n)/ (cosx_1+cosx_2+...+cosx_n) < tanx_n?

Aug 10, 2016

See explanation...

#### Explanation:

Here are the things I will assume are known:

$\tan x = \frac{\sin x}{\cos x}$

In the interval $\left(0 , \frac{\pi}{2}\right)$ we have:

$\left\{\begin{matrix}\sin x > 0 \\ \cos x > 0 \\ \tan x \text{ is strictly monotonic increasing with } x\end{matrix}\right.$

For brevity use the abbreviations:

$\left\{\begin{matrix}{s}_{k} = \sin {x}_{k} \\ {c}_{k} = \cos {x}_{k} \\ {t}_{k} = \tan {x}_{k}\end{matrix}\right.$

Note that if $a , b , c , d > 0$ then:

$\frac{a}{b} < \frac{c}{d}$ if and only if $a d < b c$

So:

${s}_{1} / {c}_{1} < {s}_{2} / {c}_{2} \text{ } \implies$

${s}_{1} {c}_{2} < {s}_{2} {c}_{1} \text{ } \implies$

${s}_{1} {c}_{2} + {s}_{1} {c}_{1} < {s}_{2} {c}_{1} + {s}_{1} {c}_{1} \text{ } \implies$

${s}_{1} \left({c}_{1} + {c}_{2}\right) < {c}_{1} \left({s}_{1} + {s}_{2}\right) \text{ } \implies$

${s}_{1} / {c}_{1} < \frac{{s}_{1} + {s}_{2}}{{c}_{1} + {c}_{2}}$

and:

${s}_{1} / {c}_{1} < {s}_{2} / {c}_{2} \text{ } \implies$

${s}_{1} {c}_{2} < {s}_{2} {c}_{1} \text{ } \implies$

${s}_{1} {c}_{2} + {s}_{2} {c}_{2} < {s}_{2} {c}_{1} + {s}_{2} {c}_{2} \text{ } \implies$

$\left({s}_{1} + {s}_{2}\right) {c}_{2} < \left({c}_{1} + {c}_{2}\right) {s}_{2} \text{ } \implies$

$\frac{{s}_{1} + {s}_{2}}{{c}_{1} + {c}_{2}} < {s}_{2} / {c}_{2}$

So we find:

${s}_{1} / {c}_{1} < \frac{{s}_{1} + {s}_{2}}{{c}_{1} + {c}_{2}} < {s}_{2} / {c}_{2}$

In general, if we have proved (for some $k < n$):

$\frac{{s}_{1} + {s}_{2} + \ldots + {s}_{k}}{{c}_{1} + {c}_{2} + \ldots + {c}_{k}} < {s}_{k} / {c}_{k}$

then since ${s}_{k} / {c}_{k} < {s}_{k + 1} / {c}_{k + 1}$ we have:

$\frac{{s}_{1} + {s}_{2} + \ldots + {s}_{k}}{{c}_{1} + {c}_{2} + \ldots + {c}_{k}} < {s}_{k + 1} / {c}_{k + 1}$

and hence:

$\frac{\left({s}_{1} + {s}_{2} + \ldots + {s}_{k}\right)}{\left({c}_{1} + {c}_{2} + \ldots + {c}_{k}\right)} < \frac{\left({s}_{1} + {s}_{2} + \ldots + {s}_{k}\right) + {s}_{k + 1}}{\left({c}_{1} + {c}_{2} + \ldots + {c}_{k}\right) + {c}_{k + 1}}$

Applying this for each $k < n$ in increasing order, we eventually get to:

${s}_{1} / {c}_{1} < \frac{{s}_{1} + {s}_{2} + \ldots + {s}_{n}}{{c}_{1} + {c}_{2} + \ldots + {c}_{n}}$

Similarly, if we have proved (for some $k > 1$):

${s}_{k} / {c}_{k} < \frac{{s}_{k} + {s}_{k + 1} + \ldots + {s}_{n}}{{c}_{k} + {c}_{k + 1} + \ldots + {c}_{n}}$

then since ${s}_{k - 1} / {c}_{k - 1} < {s}_{k} / {c}_{k}$ we have:

${s}_{k - 1} / {c}_{k - 1} < \frac{{s}_{k} + {s}_{k + 1} + \ldots + {s}_{n}}{{c}_{k} + {c}_{k + 1} + \ldots + {c}_{n}}$

and hence:

$\frac{{s}_{k - 1} + \left({s}_{k} + {s}_{k + 1} + \ldots + {s}_{n}\right)}{{c}_{k - 1} + \left({c}_{k} + {c}_{k + 1} + \ldots + {c}_{n}\right)} < \frac{{s}_{k} + {s}_{k + 1} + \ldots + {s}_{n}}{{c}_{k} + {c}_{k + 1} + \ldots + {c}_{n}}$

Applying this for each $k > 1$ in decreasing order, we eventually get to:

$\frac{{s}_{1} + {s}_{2} + \ldots + {s}_{n}}{{c}_{1} + {c}_{2} + \ldots + {c}_{n}} < {s}_{n} / {c}_{n}$

So $\tan {x}_{1} = \frac{\sin {x}_{1}}{\cos {x}_{1}} < \frac{\sin {x}_{1} + \sin {x}_{2} + \ldots + \sin {x}_{n}}{\cos {x}_{1} + \cos {x}_{2} + \ldots + \cos {x}_{n}} < \frac{\sin {x}_{n}}{\cos {x}_{n}} = \tan {x}_{n}$

Aug 10, 2016

See below

#### Explanation:

Given two real fractions such that

${x}_{1} / \left({y}_{1}\right) < {x}_{2} / \left({y}_{2}\right)$ then

${x}_{1} / \left({y}_{1}\right) < \frac{{x}_{1} + {x}_{2}}{{y}_{1} + {y}_{2}} < {x}_{2} / \left({y}_{2}\right)$ because

calling ${\lambda}_{1} = \frac{{x}_{1}}{{y}_{1}}$ and ${\lambda}_{2} = \frac{{x}_{2}}{{y}_{2}}$

we have

${\lambda}_{1} < \frac{{\lambda}_{1} {y}_{1} + {\lambda}_{2} {y}_{2}}{{y}_{1} + {y}_{2}} < {\lambda}_{2}$

but ${y}_{1} / \left({y}_{1} + {y}_{2}\right) + {y}_{2} / \left({y}_{1} + {y}_{2}\right) = 1$ so calling

$\mu = {y}_{2} / \left({y}_{1} + {y}_{2}\right)$ we have

${\lambda}_{1} < \left(1 - \mu\right) {\lambda}_{1} + \mu {\lambda}_{2} < {\lambda}_{2}$

or

${\lambda}_{1} < {\lambda}_{1} + \mu \left({\lambda}_{2} - {\lambda}_{1}\right) < {\lambda}_{2}$ This is true for $0 < \mu < 1$

Now generalizing, if we have

${x}_{1} / \left({y}_{1}\right) < {x}_{2} / \left({y}_{2}\right) < \cdots < {x}_{n} / \left({y}_{n}\right)$ then

${x}_{1} / \left({y}_{1}\right) < \frac{{\sum}_{i = 1}^{n} {x}_{i}}{{\sum}_{i = 1}^{n} {y}_{i}} < {x}_{n} / \left({y}_{n}\right)$

We know that if

$0 < {x}_{1} < {x}_{2} < \cdots < {x}_{n} < \frac{\pi}{2}$ then
$\tan \left({x}_{1}\right) < \tan \left({x}_{2}\right) < \cdots < \tan \left({x}_{n}\right)$ because

$\tan \left(x\right)$ is strict monotonic increasing in this domain