1. ## equality trigonometry

Let $\displaystyle a- b \neq k\pi$ :
$\displaystyle a\sin\alpha +b. cos\alpha = a.sin\beta +b. cos\beta =c$.

Prove: $\displaystyle cos^{2}\frac{\alpha -\beta }{2}=\frac{c^{2}}{a^{2}+b^{2}}$

2. $\displaystyle \sqrt{a^{2}+b{2}}(\frac {a}{\sqrt {a^{2}+b^{2}}}} \sin \alpha + \frac {b}{\sqrt {a^{2}+b^{2}}}} \cos \alpha) = c$

and

$\displaystyle \sqrt{a^{2}+b{2}}(\frac {a}{\sqrt {a^{2}+b^{2}}}} \sin \beta + \frac {b}{\sqrt {a^{2}+b^{2}}}} \cos \beta) = c$

let's take $\displaystyle \theta$ such that:
$\displaystyle \cos \theta = \frac {b}{\sqrt {a^{2}+b^{2}} }\quad and \quad \sin\theta = \frac {a}{\sqrt {a^{2}+b^{2}}}$
$\displaystyle \psi$ is defined as:
$\displaystyle \cos(\psi)=\frac {c}{\sqrt {a^{2}+b^{2}}}$
We have now:
$\displaystyle \cos (\alpha-\theta)=\cos(\beta-\theta)=\cos(\psi)$
Hence,
$\displaystyle 1) \alpha - \theta = \beta-\theta + 2k\pi= \psi+2k'\pi$
$\displaystyle 2) \alpha - \theta = \beta-\theta + 2k\pi= -(\psi)+2k'\pi$
$\displaystyle 3)\alpha - \theta = -(\beta-\theta) + 2k\pi= \psi+2k'\pi$
$\displaystyle 4) \alpha - \theta = -(\beta-\theta) + 2k\pi= -(\psi)+2k'\pi$

1) and 2) are not valid because they violate $\displaystyle \alpha - \beta \neq k\pi$
3) and 4) gives us:
$\displaystyle \alpha - \beta = 2 * \psi$
$\displaystyle \cos (\frac{\alpha - \beta}{2})=\cos(\psi)$
thus,
$\displaystyle \cos^{2} (\frac{\alpha - \beta}{2})= \frac{c^{2}}{a^2 + b^2}$