# Prove this inequality via graph

1. Jun 10, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
2x>3sinx-xcosx, 0<x<∏/2

2. Relevant equations

3. The attempt at a solution
One possible way is to draw the graph of the functions and compare but plotting a graph manually is not easy in this case. I want some other methods.

2. Jun 10, 2013

### SteamKing

Staff Emeritus
You know that sine and cosine vary in amplitude between 0 and 1 on the interval in question. If you substitute the end values of the interval into the inequality, that should show that the inequality does not hold.

3. Jun 10, 2013

### utkarshakash

What you are suggesting to me is just the verification of the result. Proving a result does not mean plugging the values and checking it.

4. Jun 10, 2013

### jeppetrost

Okay this is a bit messy, but it's the best I came up with. Say we define f(x)= 3sinx - xcosx, and we want to show f(x)<2x. Well, we see that at least the endpoints are all right [ f(0)=0 (okay since we don't actually consider 0), f(pi/2)=3 < pi ]. So if we can just show that the function never goes over the line 2x, we're done. Well, if we differentiate both sides, we can equivalently show that f '(x) < 2. Simply doing the differentiation gives us f '(x) = 2cosx-xsinx. We again check that the end points are okay: f '(0)= 2 (okay), f '(pi/2) = pi/2<2 (also okay). Now we need to show that f '(x) doesn't become greater than 2 (this is actually too strict, but we can anyway). So we calculate f ''(x) = x cos x-sinx and want to show that this is always negative. Well, at x=0 we have f ''(0) =0, so that's good. Now we go even further and find f '''(x) = -xsinx. This is obviously negative, which means that f ''(x) is also negative. This in turn means that f '(x) < 2, which means that f(x)<2x.

I hope you follow my train of thoughts. You just go deeper and deeper in the derivatives until you see something you know, and then go back up.

5. Jun 10, 2013

### andrien

you can just easily go with f(x)=2x-3sinx+xcosx,you can see at x=0,it is zero.so just differentiate and try to show that f'(x)>0 for the given interval.

6. Jun 10, 2013

### SteamKing

Staff Emeritus
If the inequality doesn't hold, then you are wasting time trying to prove that it does.
Verification that a relation holds at just one point in an interval is insufficient to prove that it holds at all points in the interval. However, showing that a relation fails to hold at just one point in an interval is sufficient to show that the relation does not hold.

7. Jun 10, 2013

### jeppetrost

However, it does hold, so I don't exactly see your point.

8. Jun 10, 2013

### utkarshakash

Thanks. Your method was pretty good.

9. Jun 10, 2013

### SteamKing

Staff Emeritus
Mea culpa.

10. Jun 12, 2013

### noc_njz

Why not use taylor series?