Simple trigonometric inequality, yet no simple solution.

**Doubled144314** · November 2nd 2012, 03:12 AM

Hey guys!

I've recently encountered this problem and I would like to know if any of you could present a more elegant or simple solution to it.

$cosx+tgx>sinx+1$

My solution:

When $cosx\neq0$

$$cosx+$sinx\over cosx$ $>$ $1+sinx$

$cos^2x+sinx-cosx-sinxcosx\over cosx$ $>0$

${(cosx-1)(cosx-sinx)\over cosx}>0$

${\forall_x } {cosx-1\leq 0}$ So for $cosx\neq 1$ we have:

${(cosx-sinx)\over cosx}<0$

${1-tgx}<0$

$1<tgx$ We sketch the graph, find solutions and then see what happens if $cosx=1$

**Soroban** · November 2nd 2012, 08:05 AM

Hello, Doubled144314!

I will limit my answers to Quadrant 1.
You can generalize the results if you like.

$\cos x+\tan x \:>\:\sin x+1$

We note that $x \ne \tfrac{\pi}{2}$

We have: . $\cos x + \frac{\sin x}{\cos x} \:>\:\sin x + 1$

Multiply by $\cos x\!:\;\;\cos^2\!x + \sin x \:>\:\sin x\cos x + \cos x$

. . . . $\sin x - \sin x\cos x - \cos x + \cos^2x \:>\:0$

. . . . $\sin x(1-\cos x) - \cos x(1-\cos x) \:>\:0$

. . . . . . . . . . $(1-\cos x)(\sin x - \cos x) \:>\:0$

There are two cases: . $\begin{bmatrix}\text{(1) both factors are negative} \\ \text{(2) both factors are positive}\end{bmatrix}$

Case (1): . $1-\cos x \:<\:0 \quad\Rightarrow\quad \cos x \:>\:1\;\text{ impossible}$

Case (2): . $1 - \cos \:>\:0 \quad\Rightarrow\quad \cos x\:<\:1 \quad\Rightarrow\quad x \in \left(0,\,\tfrac{\pi}{2}\right]$

. . . . . . . . $\sin x-\cos x \:>\:0 \quad\Rightarrow\quad \sin x \:>\:\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:>\:1$
. . . . . . . . $\tan x \:>\:1 \quad\Rightarrow\quad x \in \left(\tfrac{\pi}{4},\:\tfrac{\pi}{2}\right)$

Therefore: . $x \in \left(\tfrac{\pi}{4},\:\tfrac{\pi}{2}\right)$

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