Fairly simple trig question concerning calculus tools

[Asphyxiated]

Homework Statement

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

Homework Equations

Given:

[tex] Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0 [/tex]

where [tex] B \neq 0 [/tex]

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

[tex] x = X cos(\alpha) - Y sin(\alpha) [/tex]

and

[tex] y = X sin(\alpha) + Y cos(\alpha) [/tex]

and alpha is given as:

[tex] cot(2\alpha) = \frac{A-C}{B} [/tex]

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:

The Attempt at a Solution

[tex] cot(2\alpha) = \frac {A-C}{B} [/tex]

[tex] 2\alpha = cot^{-1} (\frac {A-C}{B}) [/tex]

[tex] \alpha = \frac {cot^{-1}(\frac{A-C}{B})}{2} [/tex]

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!

 

[vela]

Recall that cotangent and tangent are reciprocals.

 

[Asphyxiated]

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

[tex] \frac {cot^{-1} (\frac{A-C}{B})}{2} = \frac {tan (\frac{A-C}{B})}{2} [/tex]

right?

 

[vela]

No, you want to use the fact that

[tex]\cot 2\alpha = \frac{1}{\tan 2\alpha} = \frac{A-C}{B}[/tex]

 

[Asphyxiated]

ok so then if:

[tex] \frac {1}{tan(2\alpha)} = \frac {A-C}{B} [/tex]

then

[tex] tan(2\alpha) = \frac {B}{A-C} [/tex]

so

[tex] 2\alpha = tan^{-1} (\frac{B}{A-C}) [/tex]

[tex] \alpha = \frac {tan^{-1}(\frac{B}{A-C})}{2} [/tex]

yeah?

 

[vela]

Yup.