# Thread: Rearrange in terms of alpha

1. ## Rearrange in terms of alpha

Hi I hope I've posted this in the correct forum - I wasn't sure if it was algebra or trigonometry...

I'm trying to arrange the following equation in terms of alpha:

$\displaystyle \frac{2x - z}{2w - y} = \frac{p cos{\alpha} - 2 q cos{\beta}}{q sin{\beta} - 2 p sin{\alpha}}$

I'm just not sure where to go with it. I think there may be a trig identity I can use here but I'm not sure!

2. ## Trig Equation

Hello Frimkron

Welcome to Math Help Forum!
Originally Posted by Frimkron
Hi I hope I've posted this in the correct forum - I wasn't sure if it was algebra or trigonometry...

I'm trying to arrange the following equation in terms of alpha:

$\displaystyle \frac{2x - z}{2w - y} = \frac{p cos{\alpha} - 2 q cos{\beta}}{q sin{\beta} - 2 p sin{\alpha}}$

I'm just not sure where to go with it. I think there may be a trig identity I can use here but I'm not sure!
This is in the right place - it's definitely a trig question.

Your equation can be re-arranged into the form:

$\displaystyle a\cos\alpha +b\sin\alpha = c$, where $\displaystyle a = p(2w-y)$, etc

You then need to let $\displaystyle a\cos\alpha + b\sin\alpha = r\cos(\alpha - \theta)$

When you expand the RHS and compare coefficients of $\displaystyle \cos\alpha$ and $\displaystyle \sin\alpha$, you get:

$\displaystyle a = r\cos\theta$

$\displaystyle b = r\sin\theta$

which, if you square and add, and then divide one by the other, gives:

$\displaystyle r=\sqrt{a^2+b^2},\, \tan\theta = \frac{b}{a}$

This then gives the equation in the form

$\displaystyle r\cos(\alpha-\theta) = c$

from where you can solve for $\displaystyle \alpha$:

$\displaystyle \alpha = \arccos\Big(\frac{c}{r}\Big) + \theta$

Can you fill in the details? (It will be complicated, given your initial equation, but you'll find that's the only method that will work.)

3. Thanks Grandad this has really pointed me in the right direction!

So, rearranging I get

$\displaystyle p(2w-y)\cos\alpha+2p(2x-z)\sin\alpha = q(2x-z)\sin\beta+2q(2w-y)\cos\beta$

Then I introduce the rule

$\displaystyle a \cos\alpha + b \sin\alpha = r \cos(\alpha - \theta)$

which I assume is one of the trig identities. What is this called, by the way, and where does it come from?

Expanding the right hand side of this rule I get

$\displaystyle a \cos\alpha + b \sin\alpha = r \cos\theta \cos\alpha + r \sin\theta \sin\alpha$

therefore

$\displaystyle a = r \cos\theta$
$\displaystyle b = r \sin\theta$

and so by pythagorus

$\displaystyle r = \sqrt{a^2 + b^2}$

$\displaystyle \theta = \arctan\left(\frac{b}{a}\right)$

Rearranging $\displaystyle r \cos(\alpha-\theta) = c$ gives

$\displaystyle \alpha = \arccos\left(\frac{c}{r}\right)+\theta$

So plugging it all together I get

$\displaystyle \alpha = \arccos\left(\frac{q(2x-z)\sin\beta + 2q(2w-y)\cos\beta}{\sqrt{(p(2w-y))^2 + (2p(2x-z))^2}}\right) + \arctan\left(\frac{p(2x-z)}{2w-y}\right)$

I'm not sure how I would go about simplifying it from here. Is it likely to simplify down at all?

4. Hello Frimkron
Originally Posted by Frimkron
... Then I introduce the rule

$\displaystyle a \cos\alpha + b \sin\alpha = r \cos(\alpha - \theta)$

which I assume is one of the trig identities. What is this called, by the way, and where does it come from?...

...So plugging it all together I get

$\displaystyle \alpha = \arccos\left(\frac{q(2x-z)\sin\beta + 2q(2w-y)\cos\beta}{\sqrt{(p(2w-y))^2 + (2p(2x-z))^2}}\right) + \arctan\left(\frac{\color{red}p\color{black}(2x-z)}{2w-y}\right)$

I'm not sure how I would go about simplifying it from here. Is it likely to simplify down at all?
Your working looks fine apart from the final term, where the $\displaystyle p$ should be a $\displaystyle 2$; as in $\displaystyle ... + \arctan\left(\frac{\color{red}2\color{black}(2x-z)}{2w-y}\right)$

And you won't be able to simplify this at all, I'm afraid.

I don't think this technique has a particular name. It just uses the expansion of $\displaystyle \cos(\alpha-\theta)$ in terms of $\displaystyle \cos\alpha$ and $\displaystyle \sin\alpha$. If you like, you could have used $\displaystyle \sin(\alpha+\theta) = \sin\alpha\cos\theta +\cos\alpha\sin\theta$ instead, to get an equivalent result involving $\displaystyle \arcsin$ instead of $\displaystyle \arccos$.

If you're studying Simple Harmonic Motion (SHM) and, more especially, damped SHM, you'll find this particularly useful, since it reduces the sum of the two trig functions sine and cosine to a single one, and makes subsequent analysis - finding the amplitude and period of the oscillations, for example - much simpler.