LG a - bài 24 trang 224 sgk đại số 10 nâng cao

LG a

\(\sin (\alpha + \beta )\sin (\alpha - \beta ) \) \(= \sin ^2\alpha - {\sin ^2}\beta \)

\(={\cos ^2}\beta - {\cos ^2}\alpha \)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \sin (\alpha + \beta )\sin (\alpha - \beta )\cr& = (\sin \alpha \cos \beta + \sin \beta \cos \alpha ).\cr&(\sin \alpha \cos \beta - \sin \beta \cos \alpha ) \cr
& = {\sin ^2}\alpha {\cos ^2}\beta - {\sin ^2}\beta {\cos ^2}\alpha \cr&= {\sin ^2}\alpha (1 - {\sin ^2}\beta ) - {\sin ^2}\beta (1 - {\sin ^2}\alpha ) \cr
& = {\sin ^2}\alpha - {\sin ^2}\beta \cr &= (1 - {\cos ^2}\alpha ) - (1 - {\cos ^2}\beta ) \cr
& = {\cos ^2}\beta - {\cos ^2}\alpha \cr} \)

Chú ý: Có thể áp dụng công thức biến tích thành tổng.

LG b

\({{\tan \alpha + \tan\beta } \over {\tan \alpha - \tan\beta }} = {{\sin (\alpha + \beta )} \over {\sin (\alpha - \beta )}}\)(Khi các biểu thức có nghĩa)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \tan \alpha + \tan\beta = {{\sin \alpha } \over {\cos \alpha }} + {{\sin \beta } \over {\cos \beta }} \cr
& = {{\sin \alpha \cos \beta + \sin \beta \cos \alpha } \over {\cos \alpha \cos \beta }} \cr &= {{\sin (\alpha + \beta )} \over {\cos \alpha \cos \beta }} \cr} \)

\(\tan \alpha - \tan \beta \) \( = \dfrac{{\sin \alpha }}{{\cos \alpha }} - \dfrac{{\sin \beta }}{{\cos \beta }} \) \(= \dfrac{{\sin \alpha \cos \beta - \sin \beta \cos \alpha }}{{\cos \alpha \cos \beta }} \) \(= \dfrac{{\sin \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }}\)

Do đó: \({{\tan \alpha + \tan\beta } \over {\tan \alpha - \tan\beta }} \) \( = \dfrac{{\sin \left( {\alpha + \beta } \right)}}{{\cos \alpha \cos \beta }}:\dfrac{{\sin \left( {\alpha - \beta } \right)}}{{\cos \alpha \cos \beta }} \) \(= \dfrac{{\sin \left( {\alpha + \beta } \right)}}{{\cos \alpha \cos \beta }}.\dfrac{{\cos \alpha \cos \beta }}{{\sin \left( {\alpha - \beta } \right)}}\) \(= {{\sin (\alpha + \beta )} \over {\sin (\alpha - \beta )}}\)