What is the Range of y in (y+1)/(y-1)=sin2x/sin2a?

[zorro]

Homework Statement

I got this expression while solving a problem.

(y+1)/(y-1)=sin2x/sin2a

we need to find the range of y

Homework Equations

The Attempt at a Solution

here y=(sin2x+sin2a)/(sin2x-sin2a)

Numerator of RHS lies between sin2a-1 and 1+sin2a
Denominator lies between -1-sin2a and 1-sin2a

so RHS lies between (sin2a-1)/(-1-sin2a) and (1+sin2a)/(1-sin2a)

on solving I got y lies between tan^2 (45-a) and tan^2 (45+a)

but the answer is opposite (i.e. y does not lie between them).
Please help.

 

[HallsofIvy]

Did you remember that if a< x< b then (1/b)< 1/x< (1/a)? That, of course, is assuming that all numbers are positive. Here, since the denominator lies between -1- sin(2a) and 1- sin(2a) it can be both negative and positive and, for some x, will be 0. Of course, if the denominator goes to 0, the fraction goes to infinity so the domain can't be "between" two values.

 

[zorro]

We have to find the range of y. Not its domain.

 

[hunt_mat]

One thing that could possibly help you with this is the formulae:
$$\sin A+\sin B=2\sin\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)$$
And
$$\sin A-\sin B=2\cos\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$
The only question is now, is what is the domain of y? Once you can tell us this then we can determine the range of y

 

[vela]

Let $\beta=\sin 2x/\sin 2a$. Then

$$y=\frac{\beta+1}{\beta-1}$$

Try plotting y vs. β and then consider what the range of β is. Then perhaps you'll understand HallsofIvy's point.

 

[zorro]

Let $\beta=\sin 2x/\sin 2a$. Then

$$y=\frac{\beta+1}{\beta-1}$$

Try plotting y vs. β and then consider what the range of β is. Then perhaps you'll understand HallsofIvy's point.

Thanks, I understood this time.