# Manipulation of double-angle formula and partial derivatives.

1. Nov 30, 2011

Note: I posted this in the HW & CW section. So I think this post should be deleted here. But I can't delete it.

I'm having a difficult time trying to re-derive an already-driven result I saw in a book. I put my calculations and result, and the book's result here. I hope someone could help...

Let H be a function of a, b, tan(θ) as,

$$H^2 = 2 \{ \frac{a^2 - b^2 tan^2(\theta)}{tan^2(\theta) - 1} \}$$

and,

$$sin(2\theta) = \frac{2c^2}{a^2 + b^2}$$

Now, a, b, c, and tan(θ) depend on some a parameter m!

Now, I want to express $$\frac{\partial H^2}{\partial m}$$ explicitly as a partial derivatives of a, b, and c with respect m. $$\frac{\partial \ a^2}{\partial \ m}, \frac{\partial \ b^2}{\partial \ m}, \frac{\partial \ c^2}{\partial \ m}$$.

The book I'm reading says the result is,

$$\frac{\partial H^2}{\partial m} = \frac{2}{tan^2(\theta) - 1} \{ \frac{\partial \ a^2}{\partial \ m} - tan^2{\theta} \ \frac{\partial \ b^2}{\partial \ m} - \frac{tan(\theta)}{cos(2\theta)}[1 + \frac{H^2}{a^2 + b^2}][2 \frac{\partial \ c^2}{\partial \ m} - sin(2 \theta) (\frac{\partial \ a^2}{\partial \ m} + \frac{\partial \ b^2}{\partial \ m} ) ] \}$$

But I get a slightly different result. It's either I'm doing something wrong, or there's a trick which I'm unaware of. I would appreciate it if anyone would like to check my calculations...

I have:

$$\frac{\partial H^2}{\partial m} = \frac{2}{tan^2(\theta) - 1} [ \frac{\partial \ a^2}{\partial \ m} - tan^2(\theta) \frac{\partial \ b^2}{\partial \ m} - b^2 \frac{\partial \ tan^2(\theta)}{\partial \ m} - \frac{H^2}{2} \frac{\partial \ (tan^2(\theta) -)}{\partial \ m} ]$$

Where, I used the definition of H above.

Now, for the tan2(θ) I use:

$$\frac{\partial \ tan^2(\theta)}{\partial \ m} = \frac{\partial \ (tan^2(\theta) - 1)}{\partial \ m} = \frac{\partial \ tan^2(\theta)}{\partial \ tan(\theta)} \ \frac{\partial \ tan(\theta)}{\partial \ sin(2\theta)} \frac{\partial \ sin(2\theta)}{\partial \ m}$$

And for calculating $$\frac{\partial \ tan(\theta)}{\partial \ sin(2\theta)}$$, I express tan in terms of sin2θ as follows (maybe my mistake is here?)

$$tanθ = sin(2\theta)[1 + tan^2(\theta)]/2$$

Then I put the value of sin(2θ) in terms a,b, and c. (given at the beginning) and I get,

$$\frac{\partial H^2}{\partial m} = \frac{2}{tan^2(\theta) - 1} \{ \frac{\partial \ a^2}{\partial \ m} - tan^2{\theta} \ \frac{\partial \ b^2}{\partial \ m} - \frac{tan(\theta)}{cos(2\theta)}(1-tan^2(\theta))[\frac{b^2}{(a^2 + b^2)} + \frac{H^2}{2 (a^2 + b^2)}][2 \frac{\partial \ c^2}{\partial \ m} - sin(2 \theta) (\frac{\partial \ a^2}{\partial \ m} + \frac{\partial \ b^2}{\partial \ m} ) ] \}$$

So the difference between my result and the book's result (see above) is that I have the factor,

$$(1-tan^2(\theta))[\frac{b^2}{(a^2 + b^2)} + \frac{H^2}{2 (a^2 + b^2)}]$$

Whereas, in the book it's,

$$[1 + \frac{H^2}{a^2 + b^2}]$$

Can anyone spot my mistake?

Last edited: Nov 30, 2011