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Chứng minh rằng
a) \(\sin ({270^0} – \alpha ) = – c{\rm{os}}\alpha \)
b) \({\rm{cos}}({270^0} – \alpha ) = – \sin \alpha \)
c) \(\sin ({270^0} + \alpha ) = – c{\rm{os}}\alpha \)
d) \({\rm{cos}}({270^0} + \alpha ) = \sin \alpha \)
Gợi ý làm bài
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a) \(\eqalign{
& \sin ({270^0} – \alpha ) = \sin ({360^0} – ({90^0} + \alpha ) \cr
& = – sin({90^0} + \alpha ) = – c{\rm{os}}\alpha \cr}\)
b) \(\eqalign{
& \cos ({270^0} – \alpha ) = \cos ({360^0} – ({90^0} + \alpha )) \cr
& = \cos ({90^0} + \alpha ) = – {\rm{sin}}\alpha \cr} \)
c) \(\eqalign{
& \sin ({270^0} + \alpha ) = \sin ({360^0} – ({90^0} – \alpha )) \cr
& = – \sin ({90^0} – \alpha ) = – c{\rm{os}}\alpha \cr} \)
d) \(\eqalign{
& {\rm{cos}}({270^0} + \alpha ) = \cos ({360^0} – ({90^0} – \alpha ) \cr
& = cos({90^0} – \alpha ) = \sin \alpha \cr} \)