Simple trigonometric inequality, yet no simple solution.
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.
$\displaystyle $$cosx+tgx>sinx+1$$$
My solution:
When $\displaystyle $cosx\neq0$$
$\displaystyle $cosx+$sinx\over cosx$$\displaystyle >$$\displaystyle $1+sinx$$
$\displaystyle cos^2x+sinx-cosx-sinxcosx\over cosx$$\displaystyle >0$
$\displaystyle {(cosx-1)(cosx-sinx)\over cosx}>0$
$\displaystyle {\forall_x } {cosx-1\leq 0}$ So for $\displaystyle cosx\neq 1$ we have:
$\displaystyle {(cosx-sinx)\over cosx}<0$
$\displaystyle {1-tgx}<0$
$\displaystyle 1<tgx$ We sketch the graph, find solutions and then see what happens if $\displaystyle cosx=1$
Re: Simple trigonometric inequality, yet no simple solution.
Hello, Doubled144314!
I will limit my answers to Quadrant 1.
You can generalize the results if you like.
Quote:
$\displaystyle \cos x+\tan x \:>\:\sin x+1$
We note that $\displaystyle x \ne \tfrac{\pi}{2}$
We have: .$\displaystyle \cos x + \frac{\sin x}{\cos x} \:>\:\sin x + 1$
Multiply by $\displaystyle \cos x\!:\;\;\cos^2\!x + \sin x \:>\:\sin x\cos x + \cos x$
. . . . $\displaystyle \sin x - \sin x\cos x - \cos x + \cos^2x \:>\:0$
. . . . $\displaystyle \sin x(1-\cos x) - \cos x(1-\cos x) \:>\:0$
. . . . . . . . . . $\displaystyle (1-\cos x)(\sin x - \cos x) \:>\:0$
There are two cases: .$\displaystyle \begin{bmatrix}\text{(1) both factors are negative} \\ \text{(2) both factors are positive}\end{bmatrix}$
Case (1): .$\displaystyle 1-\cos x \:<\:0 \quad\Rightarrow\quad \cos x \:>\:1\;\text{ impossible}$
Case (2): .$\displaystyle 1 - \cos \:>\:0 \quad\Rightarrow\quad \cos x\:<\:1 \quad\Rightarrow\quad x \in \left(0,\,\tfrac{\pi}{2}\right] $
. . . . . . . . $\displaystyle \sin x-\cos x \:>\:0 \quad\Rightarrow\quad \sin x \:>\:\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:>\:1$
. . . . . . . . $\displaystyle \tan x \:>\:1 \quad\Rightarrow\quad x \in \left(\tfrac{\pi}{4},\:\tfrac{\pi}{2}\right)$
Therefore: .$\displaystyle x \in \left(\tfrac{\pi}{4},\:\tfrac{\pi}{2}\right)$