Advertisements (Quảng cáo)
Chứng minh rằng
\(\begin{array}{l}{\cos ^2}\left( {\gamma – \alpha } \right) + {\sin ^2}\left( {\gamma – \beta } \right) – 2\cos \left( {\gamma – \alpha } \right)\sin \left( {\gamma – \beta } \right)\\ = {\cos ^2}\left( {\alpha – \beta } \right)\end{array}\)
Ta có
\(\begin{array}{l}{\cos ^2}\left( {\gamma – \alpha } \right) + {\sin ^2}\left( {\gamma – \beta } \right)\\ = \dfrac{{1 + \cos 2\left( {\gamma – \alpha } \right)}}{2} + \dfrac{{1 – \cos 2\left( {\gamma – \beta } \right)}}{2}\\ = 1 + \dfrac{1}{2}\left[ {\cos 2\left( {\gamma – \alpha } \right) – \cos 2\left( {\gamma – \beta } \right)} \right]\\ = 1 + \sin \left( {2\gamma – \alpha – \beta } \right)\sin \left( {\alpha – \beta } \right)\end{array}\)
Advertisements (Quảng cáo)
Từ đó
\(\begin{array}{l}{\cos ^2}\left( {\gamma – \alpha } \right) + {\sin ^2}\left( {\gamma – \beta } \right) – 2\cos \left( {\gamma – \alpha } \right)\sin \left( {\gamma – \beta } \right)\sin \left( {\alpha – \beta } \right)\\ = 1 + \sin \left( {2\gamma – \alpha – \beta } \right)\sin \left( {\alpha – \beta } \right) – 2\cos \left( {\gamma – \alpha } \right)\sin \left( {\gamma – \beta } \right)\sin \left( {\alpha – \beta } \right)\\ = 1 + \sin \left( {\alpha – \beta } \right)\left[ {\sin \left( {2\gamma – \alpha – \beta } \right) – 2\cos \left( {\gamma – \alpha } \right)\sin \left( {\gamma – \beta } \right)} \right]\\ = 1 + \sin \left( {\alpha – \beta } \right)\left[ {\sin \left( {2\gamma – \alpha – \beta } \right) – \sin \left( {2\gamma – \alpha – \beta } \right) – \sin \left( {\alpha – \beta } \right)} \right]\\ = 1 – {\sin ^2}\left( {\alpha – \beta } \right) = {\cos ^2}\left( {\alpha – \beta } \right)\end{array}\)