Câu 6.59 trang 206 SBT Đại số 10 Nâng cao

Cho \(\sin \alpha  - \cos \alpha  = m\). Hãy tính theo \(m\)

a) \(\sin \alpha \cos \alpha ;\)

b) \(\left| {\sin \alpha  + \cos \alpha } \right|;\)

c) \({\sin ^3}\alpha  - {\cos ^3}\alpha ;\)

d) \({\sin ^6}\alpha  + {\cos ^6}\alpha .\)

Giải:

Cho \(\sin \alpha  - \cos \alpha  = m\) ta có

a)

\(\begin{array}{l}\sin \alpha \cos \alpha  =  - \dfrac{1}{2}\left[ {{{\left( {\sin \alpha  - \cos \alpha } \right)}^2} - 1} \right]\\ = \dfrac{{1 - {m^2}}}{2}.\end{array}\)

b)

\(\begin{array}{l}{\left( {\sin \alpha  + \cos \alpha } \right)^2} = 1 + 2\sin \alpha \cos \alpha \\ = 1 + 1 - {m^2} = 2 - {m^2}.\end{array}\)

Từ đó \(\left| {\sin \alpha  + \cos \alpha } \right| = \sqrt {2 - {m^2}} .\)

c)

\(\begin{array}{l}{\sin ^3}\alpha  - {\cos ^3}\alpha \\ = {\left( {\sin \alpha  - \cos \alpha } \right)^3} - 3\sin \alpha \cos \alpha \left( {\sin \alpha  - \cos \alpha } \right)\\ = {m^3} + 3\left( {\dfrac{{1 - {m^2}}}{2}} \right)m\\ = \dfrac{{m\left( {3 - {m^2}} \right)}}{2}.\end{array}\)

d)

\(\begin{array}{l}{\sin ^6}\alpha  + {\cos ^6}\alpha \\ = {\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)^3} - 3{\sin ^2}\alpha {\cos ^2}\alpha \left( {{{\sin }^2}\alpha  + co{s^2}\alpha } \right)\\ = 1 - 3{\left( {\dfrac{{1 - {m^2}}}{2}} \right)^2}\\ = \dfrac{{ - 3{m^4} + 6{m^2} + 1}}{4}.\end{array}\)

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