# Range of the given expression

1. Jul 9, 2013

### Saitama

1. The problem statement, all variables and given/known data
If $\beta$ is such that $\sin 2\beta≠0$, then show that the expression $\displaystyle \frac{x^2+2x\cos\alpha+1}{x^2+2x\cos\beta+1}$ (x real) always lies between $\cos^2\alpha/\cos^2\beta$ and $\sin^2\alpha/\sin^2\beta$.

2. Relevant equations

3. The attempt at a solution
I can rewrite the given expression to
$$\frac{(x+\cos\alpha)^2+\sin^2\alpha}{(x+\cos\beta)^2+\sin^2\beta}$$
What should I do next?

2. Jul 9, 2013

### Staff: Mentor

I think this expression has a single maximum and a single minimum (unless cos(α)=cos(β), but that special case is trivial)) and it has always the same limit for x-> +- infinity. I don't know if it is possible to find those maximal/minimal points in an analytic way, but it looks like an interesting approach.

3. Jul 9, 2013

### Saitama

I can't use limits or calculus for this problem. :(

4. Jul 9, 2013

### Staff: Mentor

Hmm... but what if you know the maximum and minimum (by some "clever guessing" aka analysis) and you can show that all other values are smaller/larger?

5. Jul 10, 2013

### haruspex

Because you have to show it lies between two values of which, in general, either could be the larger, it will probably be easiest if you break it into separate cases on that basis.
Another complication is that there's no limit given on the ranges of alpha and beta. I would try to show that it suffices to prove it for those being in the range 0 to pi/2, say.
Having done that, I would try subtracting the lower of the presumptive bounds and try to show the result is positive, etc.

6. Jul 10, 2013

### Saitama

Okay so I tried this again but I couldn't use your hints. :(

Let the given expression be y. Rearranging,
$$x^2(1-y)+2x(\cos 2\alpha-y\cos 2\beta)+(1-y)=0$$
The discriminant must be positive here. Using this condition, I was able to prove that the range (y) lies between $\cos^2\alpha/\cos^2\beta$ and $\sin^2\alpha/\sin^2\beta$. Is this a good way to solve the problem?

7. Jul 10, 2013

### haruspex

Sounds much better than anything I thought of.