1. Rearranging an equation

Hi,
I currently have the equation:

γ tan(α) = sin(α') / (cos(α') + β)

I was using α as the subject however in order to do my calculations I need α' to be the subject.
Does anyone know how I could rearrange this to that I have one case of α' on its own?
Thanks

2. Originally Posted by Topin
Hi,
I currently have the equation:

γ tan(α) = sin(α') / (cos(α') + β)

I was using α as the subject however in order to do my calculations I need α' to be the subject.
Does anyone know how I could rearrange this to that I have one case of α' on its own?
Thanks
Ummm....

In terms of symbols, is this supposed to be
$\gamma ~ tan(\alpha) = \frac{sin(\alpha ^{\prime})}{cos(\alpha ^{\prime} ) + \beta}$

Anyway:
$\gamma ~ tan(\alpha) ~ ( cos(\alpha ^{\prime} ) + \beta ) = sin(\alpha ^{\prime})$

$\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ) = sin(\alpha ^{\prime})$

Because we don't know which quadrant $\alpha ^{\prime}$ is in we need to do the following:
$(\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = sin^2(\alpha ^{\prime})$

And $sin^2( \alpha ^{\prime} ) = 1 - cos^2( \alpha ^{\prime} )$, so
$(\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = 1 - cos^2( \alpha ^{\prime} )$

When you simplify this, you will have a quadratic equation in $cos( \alpha ^{\prime} )$ which you can do the quadratic equation on.

-Dan

$(\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta ))^2 = 1 - cos^2( \alpha ^{\prime} )$
I think there was a missing bracket but I tried to rearrange this but I couldn't get all of the $cos(\alpha ^{\prime} )$ together in a way that I could then simplify. The last line of my working before I seemed to double back on myself was:
$cos^2(\alpha ^{\prime} ) + 2 cos(\alpha ^{\prime} ) \beta + \beta^2 = \frac{1 - cos^2(\alpha ^{\prime} )} {\gamma^2 tan^2 (\alpha)}$
I was wondering where I could procede from here. Thanks for your help so far though.

4. Originally Posted by topsquark
$(\gamma ~ tan(\alpha) ~ cos(\alpha ^{\prime} ) + \gamma \beta )^2 = 1 - cos^2( \alpha ^{\prime} )$
$\gamma ^2 ~ tan^2(\alpha) ~ cos^2(\alpha ^{\prime} ) + 2 \gamma~tan(\alpha)~cos(\alpha ^{\prime} ) \cdot \gamma \beta + \gamma^2 \beta^2 = 1 - cos^2( \alpha ^{\prime} )$

$(\gamma ^2 ~ tan^2(\alpha) + 1) cos^2(\alpha ^{\prime} ) + 2 \gamma^2 \beta~tan(\alpha)~cos(\alpha ^{\prime} ) + \gamma^2 \beta^2 - 1 = 0$

Set $y = cos(\alpha ^{\prime} )$.

Then we have
$(\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + 2 \gamma^2 \beta~tan(\alpha)~y + (\gamma^2 \beta^2 - 1) = 0$
which is a quadratic in y.

-Dan

5. Finally done it now. I found that in my example it came to
$(\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + (2 \gamma^2 \beta~tan^2(\alpha))~y + (\gamma^2 \beta^2 ~ tan^2 - 1) = 0
$

Using that method I've got the answer that I was expecting, also got another answer come out that I thought I would have to do another calculation for
Thankyou very much for your help - would never have managed this on my own!

6. Originally Posted by Topin
Finally done it now. I found that in my example it came to
$(\gamma ^2 ~ tan^2(\alpha) + 1) y^2 + (2 \gamma^2 \beta~tan^2(\alpha))~y + (\gamma^2 \beta^2 ~ tan^2 - 1) = 0
$

Using that method I've got the answer that I was expecting, also got another answer come out that I thought I would have to do another calculation for
Thankyou very much for your help - would never have managed this on my own!
It's a useful trick.

-Dan