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