Bài 6.30 trang 196 SBT đại số 10

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

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) \(\eqalign{
& {\rm{cos}}({270^0} + \alpha ) = \cos ({360^0} - ({90^0} - \alpha ) \cr
& = cos({90^0} - \alpha ) = \sin \alpha \cr} \)