# Fairly simple trig question concerning calculus tools

1. Jul 18, 2010

### Asphyxiated

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

Well this is technically from a calculus problem but my question focuses only on the trig of the problem so I am posting it here. This is for graphing second degree equations with a nonzero xy

2. Relevant equations
Given:

$$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$$

where $$B \neq 0$$

Use the rotation of axes equations to find an equation where B=0. Equations to do so:

$$x = X cos(\alpha) - Y sin(\alpha)$$

and

$$y = X sin(\alpha) + Y cos(\alpha)$$

and alpha is given as:

$$cot(2\alpha) = \frac{A-C}{B}$$

SO finally my question, how to solve for alpha, I think that I have just forgotten my trig or something here but an attempt I made looks like so:

3. The attempt at a solution

$$cot(2\alpha) = \frac {A-C}{B}$$

$$2\alpha = cot^{-1} (\frac {A-C}{B})$$

$$\alpha = \frac {cot^{-1}(\frac{A-C}{B})}{2}$$

and if that is correct that is all good and all but I don't remember how to solve for an inverse cotangent or how to enter it into a graphing calc so if I am right with my equation above then can someone re-enlighten me on this?

Thanks!

2. Jul 18, 2010

### vela

Staff Emeritus
Recall that cotangent and tangent are reciprocals.

3. Jul 18, 2010

### Asphyxiated

ha right ok so just so I know that I am right here

$$\frac {cot^{-1} (\frac{A-C}{B})}{2} = \frac {tan (\frac{A-C}{B})}{2}$$

right?

4. Jul 18, 2010

### vela

Staff Emeritus
No, you want to use the fact that

$$\cot 2\alpha = \frac{1}{\tan 2\alpha} = \frac{A-C}{B}$$

5. Jul 18, 2010

### Asphyxiated

ok so then if:

$$\frac {1}{tan(2\alpha)} = \frac {A-C}{B}$$

then

$$tan(2\alpha) = \frac {B}{A-C}$$

so

$$2\alpha = tan^{-1} (\frac{B}{A-C})$$

$$\alpha = \frac {tan^{-1}(\frac{B}{A-C})}{2}$$

yeah?

6. Jul 18, 2010

### vela

Staff Emeritus
Yup.