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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) \({\sin ^6}\alpha + {\cos ^6}\alpha .\)
Cho \(\sin \alpha – \cos \alpha = m\) ta có
a)
\(\begin{array}{l}\sin \alpha \cos \alpha = – \dfrac{1}{2}\left[ {{{\left( {\sin \alpha – \cos \alpha } \right)}^2} – 1} \right]\\ = \dfrac{{1 – {m^2}}}{2}.\end{array}\)
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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)
\(\begin{array}{l}{\sin ^6}\alpha + {\cos ^6}\alpha \\ = {\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^3} – 3{\sin ^2}\alpha {\cos ^2}\alpha \left( {{{\sin }^2}\alpha + co{s^2}\alpha } \right)\\ = 1 – 3{\left( {\dfrac{{1 – {m^2}}}{2}} \right)^2}\\ = \dfrac{{ – 3{m^4} + 6{m^2} + 1}}{4}.\end{array}\)