Câu 6.66 trang 208 SBT Đại số 10 Nâng cao

Chứng minh rằng

\(\begin{array}{l}{\cos ^2}\left( {\gamma  - \alpha } \right) + {\sin ^2}\left( {\gamma  - \beta } \right) - 2\cos \left( {\gamma  - \alpha } \right)\sin \left( {\gamma  - \beta } \right)\\ = {\cos ^2}\left( {\alpha  - \beta } \right)\end{array}\)

Giải:

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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