Đề bài
Chứng minh rằng
\[\begin{array}{l}{\cos ^2}\left[ {\gamma - \alpha } \right] + {\sin ^2}\left[ {\gamma - \beta } \right] - 2\cos \left[ {\gamma - \alpha } \right]\sin \left[ {\gamma - \beta } \right]\\ = {\cos ^2}\left[ {\alpha - \beta } \right]\end{array}\]
Lời giải chi tiết
Ta có
\[\begin{array}{l}{\cos ^2}\left[ {\gamma - \alpha } \right] + {\sin ^2}\left[ {\gamma - \beta } \right]\\ = \dfrac{{1 + \cos 2\left[ {\gamma - \alpha } \right]}}{2} + \dfrac{{1 - \cos 2\left[ {\gamma - \beta } \right]}}{2}\\ = 1 + \dfrac{1}{2}\left[ {\cos 2\left[ {\gamma - \alpha } \right] - \cos 2\left[ {\gamma - \beta } \right]} \right]\\ = 1 + \sin \left[ {2\gamma - \alpha - \beta } \right]\sin \left[ {\alpha - \beta } \right]\end{array}\]
Từ đó
\[\begin{array}{l}{\cos ^2}\left[ {\gamma - \alpha } \right] + {\sin ^2}\left[ {\gamma - \beta } \right] - 2\cos \left[ {\gamma - \alpha } \right]\sin \left[ {\gamma - \beta } \right]\sin \left[ {\alpha - \beta } \right]\\ = 1 + \sin \left[ {2\gamma - \alpha - \beta } \right]\sin \left[ {\alpha - \beta } \right] - 2\cos \left[ {\gamma - \alpha } \right]\sin \left[ {\gamma - \beta } \right]\sin \left[ {\alpha - \beta } \right]\\ = 1 + \sin \left[ {\alpha - \beta } \right]\left[ {\sin \left[ {2\gamma - \alpha - \beta } \right] - 2\cos \left[ {\gamma - \alpha } \right]\sin \left[ {\gamma - \beta } \right]} \right]\\ = 1 + \sin \left[ {\alpha - \beta } \right]\left[ {\sin \left[ {2\gamma - \alpha - \beta } \right] - \sin \left[ {2\gamma - \alpha - \beta } \right] - \sin \left[ {\alpha - \beta } \right]} \right]\\ = 1 - {\sin ^2}\left[ {\alpha - \beta } \right] = {\cos ^2}\left[ {\alpha - \beta } \right]\end{array}\]