Да се докаже..

[Гост]

Докажете, че за всеки триъгълник е изпълнено:
p=4Rx(cos[tex]\varphi[/tex]/2)x(cos\beta/2)x(cos\gamma/2)

[Davids]

От синусова теорема изразяваме:
$a = 2Rsin\alpha$
$b = 2Rsin\beta$
$c = 2Rsin\gamma$

Тоест:
$p = \frac{a+b+c} {2} = R(sin\alpha + sin\beta + sin\gamma) $
$= R(sin\alpha + sin\beta + sin[180° - (\alpha + \beta)]) $
$= R(sin\alpha + sin\beta + sin(\alpha + \beta)) $
$= R(sin\alpha + sin\beta + sin\alpha cos\beta + cos\alpha sin\beta) $
$= R(sin\alpha(1 + cos\beta) + sin\beta(1 + cos\alpha)) $
$ = R(2sin\frac{\alpha}{2}cos\frac{\alpha}{2}.2cos^2\frac{\beta}{2} + 2sin\frac{\beta}{2}cos\frac{\beta}{2}.2cos^2\frac{\alpha}{2}$
$= 4Rcos\frac{\alpha}{2}cos\frac{\beta}{2}(sin\frac{\alpha}{2}cos\frac{\beta}{2} + cos\frac{\alpha}{2}sin\frac{\beta}{2}) $
$= 4Rcos\frac{\alpha}{2}cos\frac{\beta}{2}.sin(\frac{\alpha + \beta} {2})$
$= 4Rcos\frac{\alpha}{2}cos\frac{\beta}{2}cos(90° - \frac{\alpha + \beta} {2})$
$= 4Rcos\frac{\alpha}{2}cos\frac{\beta}{2}cos\frac{\gamma}{2}$

[Гост]

Благодаря!