# Thread: Amount of Rotation Formula

1. ## Amount of Rotation Formula

6x²-6xy+14y²=45

write it in rotated x´y´-system (prime or whatever)

please list it out in steps.
i understand the first step, but i do not understand how to get Θ.
cot2Θ=(A-C)/B
so: cot2Θ=(6-14)/-6 =(-8)/-6 = 4/3
so how can you do: cot2Θ=4/3 !!!!!!!

2. Originally Posted by needmathhelptoujours
6x²-6xy+14y²=45

write it in rotated x´y´-system (prime or whatever)

please list it out in steps.
i understand the first step, but i do not understand how to get Θ.
cot2Θ=(A-C)/B
so: cot2Θ=(6-14)/-6 =(-8)/-6 = 4/3
so how can you do: cot2Θ=4/3 !!!!!!!
If $\cot2\theta=4/3$ then $\tan2\theta=3/4$. (You don't have to flip the thing upside down like that. It's just that I know the formula for $\tan2\theta$ but not the formula for $\cot2\theta$.)

Write $t = \tan\theta$. Then $\tan2\theta = \frac{2t}{1-t^2} = \frac34$. That is a quadratic equation for t, with solutions t=–3, t=1/3. Taking t=1/3, you get $\sin\theta = 1/\sqrt{10}$ and $\cos\theta = 3/\sqrt{10}$, and you can plug those values into the formula for rotating axes. You could equally well take t=–3, which would have the effect of rotating the axes in the opposite direction. In either case, you should end up with an equation without an x'y' term.

3. Hello, needmathhelptoujours!

We have: . $\cot2\theta \:=\:\frac{4}{3}\:=\:\frac{adj}{opp}$

$2\theta$ is in a right triangle with: . $opp = 3,\;adj = 4$
Code:
                  *
* |
*   |
*     | 3
*       |
* 2θ      |
* - - - - - *
4
Pythagorus says: . $hyp = 5$

. . Hence: . $\cos2\theta \:=\:\frac{4}{5}$

We have these two half-angle identities:

. . $\sin\theta \:=\:\sqrt{\frac{1-\cos2\theta}{2}} \qquad\cos\theta \:=\: \sqrt{\frac{1+\cos2\theta}{2}}$

Therefore: . $\begin{Bmatrix}\;\sin\theta &=& \sqrt{\dfrac{1-\frac{4}{5}}{2}} &=& \sqrt{\dfrac{1}{10}} &=& \dfrac{1}{\sqrt{10}}\; \\ \\[-3mm]
\;\cos\theta &=& \sqrt{\dfrac{1+\frac{4}{5}}{2}} &=& \sqrt{\dfrac{9}{10}} &=& \dfrac{3}{\sqrt{10}}\; \end{Bmatrix}$