Тригонометрия

[skadevil]

Здравейте! Може ли малко помощ с тази задача:
Да се докаже, че ако [tex]\alpha,\beta,\gamma[/tex] са ъгли в триъгълник, то:
[tex]cos\alpha+cos\beta+cos\gamma = 4 sin\frac{\alpha}{2}sin\frac{\beta}{2}sin\frac{\gamma}{2}+1[/tex]

[S.B.]

Здравейте! Може ли малко помощ с тази задача:
Да се докаже, че ако [tex]\alpha,\beta,\gamma[/tex] са ъгли в триъгълник, то:
[tex]cos\alpha+cos\beta+cos\gamma = 4 sin\frac{\alpha}{2}sin\frac{\beta}{2}sin\frac{\gamma}{2}+1[/tex]

[tex]cos\alpha + cos\beta + cos\gamma =[/tex]

$= cos\alpha + 2cos\frac{\beta + \gamma}{2}.cos\frac{\beta - \gamma}{2} = $

$= cos\alpha + 2cos \frac{\beta + \pi - (\alpha + \beta)}{2}.cos\frac{\beta - \gamma}{2} = $

$= cos\alpha + 2cos\frac{\beta + \pi -\alpha - \beta}{2}.cos\frac{\beta - \gamma}{2} = $

$= cos\alpha + 2cos(\frac{\pi}{2} - \frac{\alpha}{2}).cos\frac{\beta - \gamma}{2} = $

$= cos\alpha + 2sin\frac{\alpha}{2}.cos\frac{\beta - \gamma}{2} = $

$= 1 - 1 + cos\alpha + 2sin\frac{\alpha}{2}.cos\frac{\beta - \gamma}{2} = $

$= 1 - (1 - cos\alpha) + 2sin\frac{\alpha}{2}.cos\frac{\beta - \gamma}{2} = $

$= 1 - 2sin^{2}\frac{\alpha}{2} + 2sin\frac{\alpha}{2}.cos\frac{\beta - \gamma}{2} = $

$= 1 + 2sin\frac{\alpha}{2}(cos\frac{\beta - \gamma}{2} - sin\frac{\alpha}{2}) = $
$= 1 + 2sin\frac{\alpha}{2}(cos\frac{\beta - \gamma}{2} - cos(\frac{\pi}{2} - \frac{\alpha}{2})) = $

$= 1 - 4sin\frac{\alpha}{2}.sin\frac{\beta - \gamma + \pi - \alpha}{4}.sin\frac{\beta - \gamma - \pi + \alpha}{4} = $

$= 1 - 4sin\frac{\alpha}{2}.sin\frac{\beta + (\pi - (\alpha + \gamma))}{4}.sin\frac{- \gamma -(\pi -(\alpha + \beta))}{4} = $

$= 1 - 4sin\frac{\alpha}{2}.sin\frac{2\beta}{4}.sin\frac{-2\gamma}{4} = $

$= 1 + 4sin\frac{\alpha}{2}.sin\frac{\beta}{2}.sin\frac{\gamma}{2} = $

$= 4sin\frac{\alpha}{2}.sin\frac{\beta}{2}.sin\frac{\gamma}{2} + 1 $

[skadevil]

Много ти благодаря!