# Homework Help: Derivative with a range constraint - mystified!

1. Jun 11, 2010

### Gekko

1. The problem statement, all variables and given/known data

x=sin(a) / cos(b)

a+b < pi/2
a>0, b>0
0<x<1

show that

da/dx = cos^3(b)cos(a) / cos(a+b)cos(a-b)

3. The attempt at a solution

dx/da = cos(a) / cos(b) therefore
da/dx = cos(b) / cos(a)

=cos^3(b)cos(a) / cos^2(b)cos^2(a)

However the denominator of the desired format = cos(a+b)cos(a-b) = cos^2(a)cos^2(b)-sin^2(a)sin^2(b)

Not sure how to get rid of the sin^2(a)sin^2(b) term.

Is the question wrong or is there something special that needs to be done to take into account the range of a, b and x?

Very much appreciate help as Ive been totally stumped on this and the equality is required further on

2. Jun 11, 2010

### LCKurtz

If you add and subtract cos2(a)sin2(b) in the denominator your equality comes out:

$$\frac{\cos(b)}{\cos(a)}=\frac{\cos^3(b)\cos(a)}{\cos^2(a)-\sin^2(b)}$$

At least I think it does; you can check it. The problem is if you let a = b = π/6, you get the left side = 0.9742785794, the right side = 1, and x = 0.5773502695.

So it would appear something is amiss.

3. Jun 12, 2010

### Staff: Mentor

This seems to be the same question you posted in this thread.
This equation is not an identity
$$\frac{cos(b)}{cos(a)}=\frac{cos^3(b)cos(a)}{cos^2(a)-sin^2(b)}$$
LCKurtz points out a counterexample. Another is a = π/6 and b = π/4. Using these numbers the left side value is sqrt(6)/3 and the right side value is sqrt(6)/2.