Câu 6.51 trang 205 Sách BT Đại số 10 Nâng cao:...

a) Chứng minh rằng với mọi $\alpha ,\beta$, ta có:

${\sin ^2}\left( {\alpha  + \beta } \right) = {\sin ^2}\alpha  + {\sin ^2}\beta  + 2\sin \alpha \sin \beta \cos \left( {\alpha  + \beta } \right)$

b) Biết $\cos \alpha  + \cos \beta  = m;\sin \alpha  + \sin \beta  = n,$hãy tính $\cos \left( {\alpha  - \beta } \right)$ theo m, n

c) Biết ${\cos ^2}\alpha  + {\cos ^2}\beta  = p.$ Hãy tính $\cos \left( {\alpha  - \beta } \right)\cos \left( {\alpha  + \beta } \right)$ theo p.

a)

 $\begin{array}{l}{\sin ^2}\left( {\alpha  + \beta } \right) = {\left( {\sin \alpha \cos \beta  + \sin \beta \cos \alpha } \right)^2}\\ = {\sin ^2}\alpha {\cos ^2}\beta  + {\sin ^2}\beta {\cos ^2}\alpha  + 2\sin \alpha \cos \alpha sin\beta cos\beta \\ = {\sin ^2}\alpha \left( {1 - {{\sin }^2}\beta } \right) + {\sin ^2}\beta \left( {1 - {{\sin }^2}\alpha } \right) + 2\sin \alpha \cos \alpha \sin \beta \cos \beta \\ = {\sin ^2}\alpha  + {\sin ^2}\beta  - 2{\sin ^2}\alpha {\sin ^2}\beta  + 2\sin \alpha \cos \alpha \sin \beta \cos \beta \\ = {\sin ^2}\alpha  + {\sin ^2}\beta  + 2\sin \alpha \sin \beta \left( {\cos \alpha \cos \beta  - \sin \alpha \sin \beta } \right)\\ = {\sin ^2}\alpha  + {\sin ^2}\beta  + 2\sin \alpha \sin \beta \cos \left( {\alpha  + \beta } \right)\end{array}$ 

b)

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

Do đó $\cos \left( {\alpha  - \beta } \right) = \dfrac{{{m^2} + {n^2} - 2}}{2}.$

c)

$\begin{array}{l}\cos \left( {\alpha  - \beta } \right)\cos \left( {\alpha  + \beta } \right)\\ = \dfrac{1}{2}\left( {\cos 2\alpha  + \cos 2\beta } \right)\\ = \dfrac{1}{2}\left( {2{{\cos }^2}\alpha  - 1 + 2{{\cos }^2}\beta  - 1} \right)\\ = {\cos ^2}\alpha  + {\cos ^2}\beta  - 1 = p - 1\end{array}$