Bài 34 trang 207 SGK Đại số 10 Nâng cao

Chứng minh rằng:

LG a

\({{1 - 2\sin \alpha \,\cos \alpha } \over {{{\cos }^2}\alpha  - {{\sin }^2}\alpha }} = {{1 - \tan \alpha } \over {1 + \tan \alpha }}\) khi các biểu thức đó có nghĩa

Lời giải chi tiết:

Ta có:

\(\eqalign{
& {{1 - 2\sin \alpha \,\cos \alpha } \over {{{\cos }^2}\alpha - {{\sin }^2}\alpha }} \cr & = \frac{{{{\cos }^2}\alpha  + {{\sin }^2}\alpha  - 2\sin \alpha \cos \alpha }}{{{{\cos }^2}\alpha  - {{\sin }^2}\alpha }}\cr &= {{{{(cos\alpha - \sin \alpha )}^2}} \over {(cos\alpha - \sin \alpha )(cos\alpha + \sin \alpha )}} \cr 
& = {{(cos\alpha - \sin \alpha )} \over {(cos\alpha + \sin \alpha )}} \cr& = \frac{{\cos \alpha \left( {1 - \frac{{\sin \alpha }}{{\cos \alpha }}} \right)}}{{\cos \alpha \left( {1 + \frac{{\sin \alpha }}{{\cos \alpha }}} \right)}}\cr&= {{\cos \alpha (1 - \tan \alpha )} \over {\cos \alpha (1 + tan\alpha )}} \cr 
& = {{1 - \tan \alpha } \over {1 + \tan \alpha }} \cr} \) 

LG b

\(ta{n^2}\alpha {\rm{ }} - {\rm{ }}si{n^2}\alpha {\rm{ }} = {\rm{ }}ta{n^2}\alpha {\rm{ }}si{n^2}\alpha \)

Lời giải chi tiết:

Ta có: \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} \) \(\Rightarrow \sin \alpha  = \tan \alpha \cos \alpha \) \(  \Rightarrow {\sin ^2}\alpha  = {\tan ^2}\alpha {\cos ^2}\alpha \)

Do đó:

\(ta{n^2}\alpha {\rm{ }} - {\rm{ }}si{n^2}\alpha {\rm{ }} \) \(= {\tan ^2}\alpha  - {\tan ^2}\alpha .{\cos ^2}\alpha \) \(= {\rm{ }}ta{n^2}\alpha ({\rm{ }}1 - {\rm{ }}co{s^2}\alpha ){\rm{ }} \) \(= {\rm{ }}ta{n^2}\alpha {\rm{ }}si{n^2}\alpha \)

LG c

\(2(1{\rm{ }}-\sin\alpha {\rm{ }})\left( {1{\rm{ }} + {\rm{ }}cos\alpha } \right){\rm{ }} \) \(= {\rm{ }}{\left( {1{\rm{ }} - {\rm{ }}\sin\alpha {\rm{ }} + {\rm{ }}\cos\alpha {\rm{ }}} \right)^2}\)

Lời giải chi tiết:

Ta có:

\(VT=2(1-si{n}\alpha {\rm{ }})\left( {1{\rm{ }} + {\rm{ }}cos\alpha } \right){\rm{ }}\)

\( = 2\left( {1 - \sin \alpha  + \cos \alpha  - \sin \alpha \cos \alpha } \right)\)

\(= {\rm{ }}2{\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }} + {\rm{ }}2cos\alpha {\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }}cos\alpha \)

\(\begin{array}{l}
VP = {\left( {1 - \sin \alpha + \cos \alpha } \right)^2}\\
= 1 + {\sin ^2}\alpha + {\cos ^2}\alpha \\
- 2\sin \alpha + 2\cos \alpha - 2\sin \alpha \cos \alpha \\
= 2 - 2\sin \alpha + 2\cos \alpha - 2\sin \alpha \cos \alpha \\
\Rightarrow VT = VP \Rightarrow dpcm
\end{array}\)