Proving a Trigonometric Inequality: Cosθ < (Sinθ)/θ < 1/Cosθ

[isalloum4]

How to prove:

Cosθ < (Sinθ)/θ < 1/Cosθ when 0<x<1/2π : π is Pie

It seems to be a fundamental inequality that Apostol calculus uses in its text.

Thank you

 

[Dick]

How to prove:

Cosθ < (Sinθ)/θ < 1/Cosθ when 0<x<1/2π : π is Pie

It seems to be a fundamental inequality that Apostol calculus uses in its text.

Thank you

Depends on what you are allowed to use to prove it. The first half of the inequality is basically the same thing as showing x<tan(x). The second is the same as showing sin(2x)<2x. Can you show those?

 

[isalloum4]

I don't know how to show these. I would appreciate it if you show me how or tell me about a source that could help me with this.
Much appreciated.

 

[Dick]

I don't know how to show these. I would appreciate it if you show me how or tell me about a source that could help me with this.
Much appreciated.

I don't have Apostol, so I don't know what you are allowed to use here. Both are easy using derivatives and the mean value theorem, i.e. calculus. Have you gotten that far?

 

[isalloum4]

Actually I am self teaching myself and I just started with Apostol. Apostol starts with intergration first. So I didn't get to derivation yet. Maybe I have to wait.
Thank you for your time.

 

[Dick]

Actually I am self teaching myself and I just started with Apostol. Apostol starts with intergration first. So I didn't get to derivation yet. Maybe I have to wait.
Thank you for your time.

That's an interesting approach to teaching calculus. I've never really seen it done that way. In that case, maybe you do have to just assume it for now and wait for the proof. Or maybe somebody knows a way to do it without derivatives. Good for you for putting in the effort to self-study! Good question. Though thinking about it, it's pretty easy to show sin(x)<x using the geometric meaning of x and sin(x) in terms of distances on the unit circle. Probably ditto for x<tan(x). Apostol doesn't give a proof?

 

[lurflurf]

The usual geometric proof is to consider a triangle with sides (sin x, cos x, 1) a triangle with sides (tan x,1,sec x) and a circular sector with sides (1,x/2,1). Since they can be nested we have
2A1<2A2<2A3
sin x cos x<x<tan x
so
sin x/(2A3)<sin x/(2A2)<sin x/(2A1)
cos x<sin x/x<sec x

If I recall correctly Apostol defines 2sin x as the chord of a sector of area x (or something close to that). So this is a natural result.

 

[voko]

Apostol defines the sine and cosine functions via a few axioms, one of those is the inequality in question. A few sections later, he re-defines them geometrically and proves that the axioms are satisfied. So the proof is in the book, just stay with it.

 

[HallsofIvy]

How you prove that depends, as others have said, upon how you define sine and cosine. A common definition, in Calculus and PreCalculus texts is to define them in terms of the "unit circle"- the circle of radius 1 with center at the origin of a coordinate system. Given such a circle, to find cos(t) and sin(t), start at (1, 0) and measure around the circle (counter-clockwise if t> 0, clockwise if t< 0) a distance |t|. cos(t) is the x-coordinate of the final point, sin(t) is the y-coordinate of that point.

Now, draw a line from (cos(t), sin(t)) to the x-axis, a straight line from (cos(t), sin(t)) to (0, 0).
Those two lines, together with the x-axis itself, form a right triangle having legs of length y and x and so area of (1/2)sin(t)cos(t).
The "sector" having as boundaries the x-axis, the line from (cos(t), sin(t)) to (0, 0) and the arc of the circle, has area $t/2$ (the area of a circular sector with radius r and angle $\theta$ has area $\theta r/2$. Here, r= 1 and $\theta= t$.)

Draw a vertical line at (1, 0) (tangent to the circle). That intersects the line from (cos(t) sin(t)) to (0, 0) at (1, tan(t)).
(The line from (cos(t), sin(t)) to (0, 0) has slope (sin(t)- 0)/(cos(t)- 0)= sin(t)/cos(t)= tan(t) and so has equation y= tan(t) x. Setting x= 1, y= tan(t).)
So this gives a right triangle with legs of length 1 and tan(t) and so area (1/2)(1)(tan(t)= (1/2)tan(t).

Now, clearly, the first right triangle is inside the circular sector which is inside the last right triangle. That is the area of each is less than the area of the next:
$$(1/2)sin(t)cos(t)\le (1/2)t\le (1/2)tan(t)$$
And obvious thing to do is multiply through by 2:
$$sin(t)cos(t)\le t\le tan(t)= \frac{sin(t)}{cos(t)}$$
Divide each part by sin(t). (We are taking the limit as t goes to 0 so t is NOT itself 0 and sin(t) is not 0)
$$cos(t)\le \frac{t}{sin(t)}\le \frac{1}{cos(t)}$$
Taking the reciprocal of each part reverses the inequalities:
$$\frac{1}{cos(t)}\ge \frac{sin(t)}{t}\ge cos(t)$$
or
$$cos(t)\le \frac{sin(t)}{t}\le \frac{1}{cos(t)}$$

That is the result you wanted. The application of that, of course, is to observe that cos(0) is 1 and cosine is a continuous function to taking the limit as t goes to 0, both left and right parts go to 0:
$$1\le \lim_{t\to 0}\frac{sin(t)}{t}\le 1$$
which proves that $\lim_{t\to 0} sin(t)/t= 1$.