Question

Докажите ,что если alpha ,beta ,gamma - углы треугольника ,то выполняется равенство :
sin^2 alfa + sin^2 betta + sin^2 gamma = 2 + 2cos alpha cos beta cos gamma

Answer 1

Объяснение:

${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma = 2 + 2\cos \alpha \cos \beta \cos \gamma$

$\gamma = \pi - (\alpha + \beta );\,\,\alpha + \beta = \pi - \gamma$

${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma = {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}(\pi - (\alpha + \beta )) =$

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

$=\frac{3}{2} - \frac{1}{2}(\cos 2\alpha + \cos 2\beta + \cos 2(\alpha + \beta )) =$

$=\frac{3}{2} - \frac{1}{2}(2\cos (\alpha + \beta )\cos (\alpha - \beta ) + 2{\cos ^2}(\alpha + \beta ) - 1) =$

$=2 - \cos (\alpha + \beta )\cos (\alpha - \beta ) - {\cos ^2}(\alpha + \beta ) = 2 - \cos (\alpha + \beta )(\cos (\alpha - \beta ) + \cos (\alpha + \beta )) =$

$=2 - \cos (\pi - \gamma ) \cdot 2\cos \alpha \cos \beta = 2 + 2\cos \alpha \cos \beta \cos \gamma .$