Đề bài - câu 39 trang 243 sbt đại số 10 nâng cao
|
\(\begin{array}{l}{\cos ^2}\gamma + 2\cos \alpha \cos \beta \cos \gamma \\ = \cos \gamma \left[ {\cos \left( {\pi - \left( {\alpha + \beta } \right)} \right) + 2\cos \alpha \cos \beta } \right]\\ = \cos \gamma \left[ { - \cos \alpha \cos \beta + \sin \alpha \sin \beta + 2\cos \alpha \cos \beta } \right]\\ = \cos \gamma \cos \left( {\alpha - \beta } \right)\\ = - \cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right)\\ = {\sin ^2}\alpha {\sin ^2}\beta - {\cos ^2}\alpha {\cos ^2}\beta \\ = {\sin ^2}\alpha {\sin ^2}\beta - \left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\sin }^2}\beta } \right)\\ = - 1 + {\sin ^2}\alpha + {\sin ^2}\beta \\ = 1 - {\cos ^2}\alpha - {\cos ^2}\beta .\end{array}\) Đề bài Chứng minh rằng, nếu \(\alpha + \beta + \gamma = \pi \) thì \({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma + 2\cos \alpha \cos \beta \cos \gamma = 1\). Lời giải chi tiết Ta có: \(\begin{array}{l}{\cos ^2}\gamma + 2\cos \alpha \cos \beta \cos \gamma \\ = \cos \gamma \left[ {\cos \left( {\pi - \left( {\alpha + \beta } \right)} \right) + 2\cos \alpha \cos \beta } \right]\\ = \cos \gamma \left[ { - \cos \alpha \cos \beta + \sin \alpha \sin \beta + 2\cos \alpha \cos \beta } \right]\\ = \cos \gamma \cos \left( {\alpha - \beta } \right)\\ = - \cos \left( {\alpha + \beta } \right)\cos \left( {\alpha - \beta } \right)\\ = {\sin ^2}\alpha {\sin ^2}\beta - {\cos ^2}\alpha {\cos ^2}\beta \\ = {\sin ^2}\alpha {\sin ^2}\beta - \left( {1 - {{\sin }^2}\alpha } \right)\left( {1 - {{\sin }^2}\beta } \right)\\ = - 1 + {\sin ^2}\alpha + {\sin ^2}\beta \\ = 1 - {\cos ^2}\alpha - {\cos ^2}\beta .\end{array}\)
|
