Bài tập 13 trang 11 SBT Toán 11 Tập 1 Cánh diều - CD

a) Ta có: \({\left( {\sin \alpha + \cos \alpha } \right)^2}\)

\(= {\sin ^2}\alpha + 2\sin \alpha .\cos \alpha + {\cos ^2}\alpha \\= 1 + 2\sin \alpha \cos \alpha \)

Suy ra \(A = \sin \alpha .\cos \alpha = \frac{{{{\left( {\sin \alpha + \cos \alpha } \right)}^2} - 1}}{2}= \frac{{{{\left( {\frac{1}{3}} \right)}^2} - 1}}{2} = - \frac{4}{9}\)

b) Ta có: \({B^2} = {\left( {\sin \alpha - \cos \alpha } \right)^2}\)

\(= {\sin ^2}\alpha - 2\sin \alpha .\cos \alpha + {\cos ^2}\alpha \\= 1 - 2\sin \alpha \cos \alpha \)

Theo câu a, ta có \(\sin \alpha .\cos \alpha = - \frac{4}{9}\).

Nên \({B^2} = 1 - 2\left( { - \frac{4}{9}} \right) = \frac{{17}}{9} \Rightarrow B = \pm \frac{{\sqrt {17} }}{3}\).

Do \( - \frac{\pi }{2} < \alpha < 0\) , ta suy ra \(\sin \alpha < 0\), \(\cos \alpha > 0\).

Từ đó \(B = \sin \alpha - \cos \alpha < 0\).

Như vậy \(B = - \frac{{\sqrt {17} }}{3}\).

c) Ta có: \({\left( {\sin \alpha + \cos \alpha } \right)^3} = {\sin ^3}\alpha + {\cos ^3}\alpha + 3\sin \alpha .\cos \alpha \left( {\sin \alpha + \cos \alpha } \right)\)

Theo câu a, ta có \(\sin \alpha .\cos \alpha = - \frac{4}{9}\) nên:

\(C = {\left( {\sin \alpha + \cos \alpha } \right)^3} - 3\sin \alpha .\cos \alpha \left( {\sin \alpha + \cos \alpha } \right) = {\left( {\frac{1}{3}} \right)^3} - 3.\frac{{ - 4}}{9}.\frac{1}{3} = \frac{{13}}{{27}}\)

d) Ta có: \({\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)^2}\)

\(= {\left( {{{\sin }^2}\alpha } \right)^2} + {\left( {{{\cos }^2}\alpha } \right)^2} + 2{\sin ^2}\alpha {\cos ^2}\alpha \)

\( = {\sin ^4}\alpha + {\cos ^4}\alpha + 2{\sin ^2}\alpha {\cos ^2}\alpha \)

-- Mod Toán 11 HỌC247