Đề bài - câu 6.66 trang 208 sbt đại số 10 nâng cao

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

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

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

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