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?