Solving a Trig Inequality: A Graphical Approach

[ehrenfest]

[SOLVED] trig inequality

Homework Statement

Use the graph of y = sin x to show the following. Given a triangle ABC,

$$\frac{\sin B + \sin C}{2}\leq \sin \frac{B+C}{2}$$.

Homework Equations

The Attempt at a Solution

Does anyone else think this question is weird? B < 180 - C so shouldn't their be a strict inequality? I am really not sure how to use the graphs. Should I replace B with 180 - C and try to derive some new inequalities? Can I use that inequality that argument of the sine to get an inequality with the sine itself?

 

[dynamicsolo]

Pick two points B and C in the interval (0, pi). On the graph,

$$\frac{\sin B + \sin C}{2}$$ would be at the midpoint of the line segment connecting (B, sin B) with (C, sin C), which is at x = (B+C)/2 , whereas

$$\sin \frac{B+C}{2}$$ is on the curve y = sin x at x = (B+C)/2 , the midpoint of the interval [B,C].

The curve is above the line segment unless B = C.

Just out of curiosity, what did they want this inequality for?

 

[ehrenfest]

Pick two points B and C in the interval (0, pi).

So, this actually proves something more general than the case when B and C are angles in a triangle, right? You're proof would hold if B = C = 2pi/3.

I think I got thrown off thinking about how to implement the restriction that A and B were angles in a triangle when it was easier to just prove this more general result.

Just out of curiosity, what did they want this inequality for?

Its an exercise from Loren Larsen's book Problem Solving Through Problems.

 

[dynamicsolo]

So, this actually proves something more general than the case when B and C are angles in a triangle, right? You're proof would hold if B = C = 2pi/3.

Yeah, I guess it is more general. It struck me that B and C wouldn't go just anywhere in the interval if they represented angles in a triangle, but that didn't actually matter for the proof. The interval (0, pi) is important only in that sin x is always positive there, so it is simple to make a statement about the line segment and the curve.

Its an exercise from Loren Larsen's book Problem Solving Through Problems.

Ah, I have that book here (it's my partner's copy): I should look through it some more! The problems in books such as that often try to get you accustomed to looking for solutions in less analytical and more intuitive or heuristic ways. (Sometimes the way the problem is set provides a hint as to what you might look for.)

 

[Gib Z]

An extremely more pointless and exhausting method of proof for the inequality is to replace sin B and sin C with their series counterparts and prove the inequality algebraically. It's not fun at all, I don't recommend it. I don't even know why I bothered.

 

[dynamicsolo]

A really cheap way to prove this inequality is to use the "sum-to-product" relation

sin B + sin C = 2 · sin[(B+C)/2] · cos[(B-C)/2] ,

preceded by as much "machinery" as you feel you need to prove this. Again, the equality in the proposition you wish to prove only holds when B = C . The sign of the difference is never an issue, since cosine is an even function and the argument of the cosine term won't exceed pi in the interval of interest.

... It's not fun at all, I don't recommend it. I don't even know why I bothered.

You shouldn't feel that taking alternate approaches like that are a waste: sometimes you notice or learn something of interest for future reference (or can at least warn later students what methods don't work too well...). Sometimes you come to understand something new about some familiar topic by looking at it from an uncommon angle. I'll say, though, that I've also sometimes found that series methods can make it harder to extract certain information unless you're really familiar with expressions for all sorts of functions (I'm not saying I am; I understand Feynman kept notes on a lot of functions which turned out to be useful later...)