# equality trigonometry

• Feb 23rd 2011, 12:40 PM
arm
equality trigonometry
Let $a- b \neq k\pi$ :
$a\sin\alpha +b. cos\alpha = a.sin\beta +b. cos\beta =c$.

Prove: $cos^{2}\frac{\alpha -\beta }{2}=\frac{c^{2}}{a^{2}+b^{2}}$
• Feb 23rd 2011, 02:21 PM
mestmoha
$\sqrt{a^{2}+b{2}}(\frac {a}{\sqrt {a^{2}+b^{2}}}} \sin \alpha + \frac {b}{\sqrt {a^{2}+b^{2}}}} \cos \alpha) = c$

and

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

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

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