Chứng minh rằng với mọi \(\alpha ,\beta \), ta có:
LG a
\({\sin ^2}\left( {\alpha + \beta } \right) = {\sin ^2}\alpha + {\sin ^2}\beta + 2\sin \alpha \sin \beta \cos \left( {\alpha + \beta } \right)\)
Lời giải chi tiết:
\(\begin{array}{l}{\sin ^2}\left( {\alpha + \beta } \right) = {\left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right)^2}\\ = {\sin ^2}\alpha {\cos ^2}\beta + {\sin ^2}\beta {\cos ^2}\alpha + 2\sin \alpha \cos \alpha sin\beta cos\beta \\ = {\sin ^2}\alpha \left( {1 - {{\sin }^2}\beta } \right) + {\sin ^2}\beta \left( {1 - {{\sin }^2}\alpha } \right) + 2\sin \alpha \cos \alpha \sin \beta \cos \beta \\ = {\sin ^2}\alpha + {\sin ^2}\beta - 2{\sin ^2}\alpha {\sin ^2}\beta + 2\sin \alpha \cos \alpha \sin \beta \cos \beta \\ = {\sin ^2}\alpha + {\sin ^2}\beta + 2\sin \alpha \sin \beta \left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta } \right)\\ = {\sin ^2}\alpha + {\sin ^2}\beta + 2\sin \alpha \sin \beta \cos \left( {\alpha + \beta } \right)\end{array}\)
LG b
Biết \(\cos \alpha + \cos \beta = m;\sin \alpha + \sin \beta = n,\)hãy tính \(\cos \left( {\alpha - \beta } \right)\) theo m, n
Lời giải chi tiết:
\(\begin{array}{l}{m^2} + {n^2} = {\left( {\cos \alpha + \cos \beta } \right)^2} + {\left( {\sin \alpha + \sin \beta } \right)^2}\\ = {\cos ^2}\alpha + {\sin ^2}\alpha + {\cos ^2}\beta + {\sin ^2}\beta + 2\left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right)\\ = 2 + 2\cos \left( {\alpha - \beta } \right)\end{array}\)
Do đó \(\cos \left( {\alpha - \beta } \right) = \dfrac{{{m^2} + {n^2} - 2}}{2}.\)
LG c
Biết \({\cos ^2}\alpha + {\cos ^2}\beta = p.\) Hãy tính \(\cos \left( {\alpha - \beta } \right)\cos \left( {\alpha + \beta } \right)\) theo p.
Lời giải chi tiết:
\(\begin{array}{l}\cos \left( {\alpha - \beta } \right)\cos \left( {\alpha + \beta } \right)\\ = \dfrac{1}{2}\left( {\cos 2\alpha + \cos 2\beta } \right)\\ = \dfrac{1}{2}\left( {2{{\cos }^2}\alpha - 1 + 2{{\cos }^2}\beta - 1} \right)\\ = {\cos ^2}\alpha + {\cos ^2}\beta - 1 = p - 1\end{array}\)
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