Is there a non-geometic proof of this inequality

[Mu naught]

sinx ≤ x ≤ tanx

You need the above inequality to prove that lim x-> 0 of sinx/x = 1, but I've only ever seen it derived geometrically. Is there an analytical proof of the above inequality, from which you can continue the sinx/x proof as normal?

 

[tauon]

that limit is elegantly proved using the series definition of the sin function. no need for that inequality.

 

[HallsofIvy]

As tauon suggests, it depends upon how you define the trig functions. You can define sine and cosine in terms of their McLaurin series (lots of fun proving periodicity from that!). You can also define "y= sin(x)" as the solution to the differential equation y"= -y with y(0)= 0, y'(0)= 1, and define "y= cos(x)" as the solution to the differential equation y"= -y with y(0)= 1, y'(0)= 0.

In either case, you don't need that inequality nor that limit to derive the derivatives of sine and cosine.

 

[Jerbearrrrrr]

(lots of fun proving periodicity from that!)
All the horrible, horrible memories. None of it even came up in the Analysis paper. If you just want the inequalities, without being too fussy about "how much we know" at this point:
Integrate cosx<1 for the first bit.
Integrate 1<(1/cos^2 x) for the second bit.

(We have to assume we know the derivatives of trig functions, and a few properties of integration. From the proofs involved, we probably know enough to do the sinx/x anyway, but...there it is if you need to check the result in an exam or something).

 

[CellCoree]

Yes, there is a non-geometric proof of this inequality. It can be derived using the properties of trigonometric functions and the fundamental theorem of calculus. Here is a brief explanation of the proof:

First, we can use the Maclaurin series expansion of sine and tangent functions to rewrite the inequality as:

sinx ≤ x ≤ x + x^3/3

Next, using the fundamental theorem of calculus, we can show that the derivative of sinx is cosx, and the derivative of x is 1. Therefore, we can write the inequality as:

cosx ≤ 1 ≤ 1 + x^2

Now, using the fact that cosx ≤ 1 for all values of x, we can simplify the inequality to:

1 ≤ 1 + x^2

which is always true.

Therefore, we have proven that sinx ≤ x ≤ tanx using an analytical approach. This proof can then be used to continue the proof of the limit of sinx/x as x approaches 0, as you have mentioned.