Câu 39 trang 243 SBT Toán Đại 10 Nâng cao :...

Chứng minh rằng, nếu $\alpha  + \beta  + \gamma  = \pi$ thì

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

Ta có:

$\begin{array}{l}{\cos ^2}\gamma  + 2\cos \alpha \cos \beta \cos \gamma \\ = \cos \gamma \left[ {\cos \left( {\pi  - \left( {\alpha  + \beta } \right)} \right) + 2\cos \alpha \cos \beta } \right]\\ = \cos \gamma \left[ { - \cos \alpha \cos \beta  + \sin \alpha \sin \beta  + 2\cos \alpha \cos \beta } \right]\\ = \cos \gamma \cos \left( {\alpha  - \beta } \right)\\ =  - \cos \left( {\alpha  + \beta } \right)\cos \left( {\alpha  - \beta } \right)\\ = {\sin ^2}\alpha {\sin ^2}\beta  - {\cos ^2}\alpha {\cos ^2}\beta \\ = {\sin ^2}\alpha {\sin ^2}\beta  - \left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\sin }^2}\beta } \right)\\ =  - 1 + {\sin ^2}\alpha  + {\sin ^2}\beta \\ = 1 - {\cos ^2}\alpha  - {\cos ^2}\beta .\end{array}$