- LG a
- LG b
- LG c
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:
\[\begin{array}{l}\cos \left[ {\alpha - \beta } \right]\cos \left[ {\alpha + \beta } \right]\\ = \dfrac{1}{2}\left[ {\cos 2\alpha + \cos 2\beta } \right]\\ = \dfrac{1}{2}\left[ {2{{\cos }^2}\alpha - 1 + 2{{\cos }^2}\beta - 1} \right]\\ = {\cos ^2}\alpha + {\cos ^2}\beta - 1 = p - 1\end{array}\]