Chứng minh rằng:
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
b) \(ta{n^2}\alpha {\rm{ }} - {\rm{ }}si{n^2}\alpha {\rm{ }} = {\rm{ }}ta{n^2}\alpha {\rm{ }}si{n^2}\alpha \)
c) \(2(1{\rm{ }}-\sin\alpha {\rm{ }})\left( {1{\rm{ }} + {\rm{ }}cos\alpha } \right){\rm{ }} = {\rm{ }}{\left( {1{\rm{ }} - {\rm{ }}\sin\alpha {\rm{ }} + {\rm{ }}\cos\alpha {\rm{ }}} \right)^2}\)
Đáp án
a) Ta có:
\(\eqalign{
& {{1 - 2\sin \alpha \,\cos \alpha } \over {{{\cos }^2}\alpha - {{\sin }^2}\alpha }} = {{{{(cos\alpha - \sin \alpha )}^2}} \over {(cos\alpha - \sin \alpha )(cos\alpha + \sin \alpha )}} \cr
& = {{(cos\alpha - \sin \alpha )} \over {(cos\alpha + \sin \alpha )}} = {{\cos \alpha (1 - \tan \alpha )} \over {\cos \alpha (1 + tan\alpha )}} \cr
& = {{1 - \tan \alpha } \over {1 + \tan \alpha }} \cr} \)
b) Ta có:
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\(ta{n^2}\alpha {\rm{ }} - {\rm{ }}si{n^2}\alpha {\rm{ }} = {\rm{ }}ta{n^2}\alpha ({\rm{ }}1 - {\rm{ }}co{s^2}\alpha ){\rm{ }} = {\rm{ }}ta{n^2}\alpha {\rm{ }}si{n^2}\alpha \)
c) Ta có:
\(2(1-si{n}\alpha {\rm{ }})\left( {1{\rm{ }} + {\rm{ }}cos\alpha } \right){\rm{ }}\)
\(= {\rm{ }}2{\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }} + {\rm{ }}2cos\alpha {\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }}cos\alpha \)
\( = {\rm{ }}1{\rm{ }} + {\rm{ }}si{n^2}\alpha {\rm{ }} + {\rm{ }}co{s^2}\alpha {\rm{ }}-{\rm{ }}2sin\alpha {\rm{ }} + {\rm{ }}2cos\alpha {\rm{ }} \)
\(- {\rm{ }}2sin\alpha {\rm{ }}cos\alpha \)
\( = {\rm{ }}{\left( {1{\rm{ }} - {\rm{ }}sin\alpha {\rm{ }} + {\rm{ }}cos\alpha {\rm{ }}} \right)^2}\)