Solving Sin and Tan Inequalities: Simple Methods for x > 0 and x in (0, Pi/2)

[wowolala]

i am trying to figure out 2 quesitons below:

(a) sin x < x, for all x > 0;
(b) x < tan x, for all x 2 (0;Pi/2).

at the first, i solved by the graph, but it seemed too tedious.

can someone tell me how to solve them in kinda simple ways

thx

 

[HallsofIvy]

i am trying to figure out 2 quesitons below:

(a) sin x < x, for all x > 0;
(b) x < tan x, for all x 2 (0;Pi/2).

at the first, i solved by the graph, but it seemed too tedious.

can someone tell me how to solve them in kinda simple ways

thx

It's not clear what the question is! If you want to prove "sin x< x for all x> 0", you can't- it's not true. If you want to determine for what values of x, x>0 sin x< x, then you first look at the equation sin(x)= x since those points will separate "<" from ">".

It's not all that tedious to graph y= sin(x)- you should know its shape already. And y= x is certainly not difficult to graph. Unfortunately, since both these equations involve a transcendental function (sin(x) and tan(x)) with a linear function (x), there is no "kinda simple" way to solve them. I would think that graphing on a graphing calculator would be the simples. Another way would be to use the "Newton-Raphson" numerical method.

 

[mathman]

i am trying to figure out 2 quesitons below:

(a) sin x < x, for all x > 0;
(b) x < tan x, for all x 2 (0;Pi/2).

at the first, i solved by the graph, but it seemed too tedious.

can someone tell me how to solve them in kinda simple ways

thx

You can use elementary calculus.

For (a) let f(x)=x-sinx, then f'(x)=1-cosx which is always non-negative. Therefore x-sinx is never decreasing and starts out =0 at x=0, increasing immediately.

For(b) let f(x)=tanx-x, then f'(x)=sec²x-1, which is also non-negative. Same argument as for (a).

 

[Integral]

Since the OP did not state what class he is taking I am going to assume that this is a precalc problem.

 

[Gib Z]

http://img510.imageshack.us/img510/2715/courantpage48extractgi6.th.jpg

I've uploaded a scanned extract of page 48 from Courant, Volume 1. As it already shows you from the diagram on the page, if $$0 < x < \pi/2$$, then $$1< \frac{x}{\sin x} < \frac{1}{\cos x}$$.

Since we are dealing with all positive quantities here, finding the reciprocal changes each inequality sign: $$1 > \frac{\sin x}{x} > \cos x$$. I am not going to go any further because I have some other work to do.

In fact I think I've already given you too much! You can do the rest =]

 

[HallsofIvy]

The problem I have is that the OP said

i am trying to figure out 2 quesitons below:

(a) sin x < x, for all x > 0;
(b) x < tan x, for all x 2 (0;Pi/2).

but there are no questions there!

 

[Gib Z]

I'm sure he just forgot the critical word, "show" or "prove" etc etc lol