Đề bài - bài 6 trang 154 sgk đại số 10

Lời giải chi tiết

Với \[\displaystyle {\pi \over 2}< a 0, \cos a < 0.\]

\[\displaystyle \begin{array}{l}
{\sin ^2}2a + {\cos ^2}2a = 1\\
\Rightarrow {\cos ^2}2a = 1 - {\sin ^2}2a\\
= 1 - {\left[ {\dfrac{5}{9}} \right]^2} = \dfrac{{56}}{{81}}\\
\Rightarrow \cos 2a = \pm \sqrt {\dfrac{{56}}{{81}}} = \pm \dfrac{{2\sqrt {14} }}{9}
\end{array}\]

Nếu \[\displaystyle \cos 2a = {{2\sqrt {14} } \over 9}\]thì

\[\displaystyle {\sin ^2}a = \frac{{1 - \cos 2a}}{2}\]

\[\displaystyle \eqalign{
& \Rightarrow \sin a = \sqrt {{{1 - \cos 2a} \over 2}} = \sqrt {{{1 - {{2\sqrt {14} } \over 9}} \over 2}} \cr&= {{\sqrt {9 - 2\sqrt {14} } } \over {3\sqrt 2 }}= {{\sqrt {{{\left[ {\sqrt 7 - \sqrt 2 } \right]}^2}} } \over {3\sqrt 2 }}\cr& = {{\sqrt 7 - \sqrt 2 } \over {3\sqrt 2 }} = {{\sqrt {14} - 2} \over 6} \cr} \]

\[\displaystyle {\cos ^2}a = \frac{{1 + \cos 2a}}{2}\] \[\displaystyle \Rightarrow\cos a = - \sqrt {{{1 + \cos 2a} \over 2}} \] \[\displaystyle = - \sqrt {\frac{{1 + \frac{{2\sqrt {14} }}{9}}}{2}} \] \[\displaystyle = - \sqrt {\frac{{9 + 2\sqrt {14} }}{{18}}}\] \[\displaystyle = - \frac{{\sqrt {{{\left[ {\sqrt 7 + \sqrt 2 } \right]}^2}} }}{{\sqrt {18} }}\] \[\displaystyle = - \frac{{\sqrt 7 + \sqrt 2 }}{{3\sqrt 2 }} \] \[\displaystyle = - \frac{{\sqrt {14} + 2}}{6}\]

Nếu \[\displaystyle \cos 2a = -{{2\sqrt {14} } \over 9}\]thì

\[\displaystyle {\sin ^2}a = \frac{{1 - \cos 2a}}{2}\]

\[\displaystyle \eqalign{
& \Rightarrow\sin a = \sqrt {{{1 - \cos 2a} \over 2}} = \sqrt {{{1 + {{2\sqrt {14} } \over 9}} \over 2}} \cr&= {{\sqrt {9 + 2\sqrt {14} } } \over {3\sqrt 2 }} = {{\sqrt {{{\left[ {\sqrt 7 + \sqrt 2 } \right]}^2}} } \over {3\sqrt 2 }}\cr& = {{\sqrt 7 + \sqrt 2 } \over {3\sqrt 2 }} = {{\sqrt {14} + 2} \over 6} \cr } \]

\[\displaystyle {\cos ^2}a = \frac{{1 + \cos 2a}}{2}\]

\[\displaystyle \Rightarrow\cos a = - \sqrt {{{1 + \cos 2a} \over 2}} \] \[\displaystyle = - \sqrt {\frac{{1 - \frac{{2\sqrt {14} }}{9}}}{2}} \] \[\displaystyle = - \sqrt {\frac{{9 - 2\sqrt {14} }}{{18}}}\] \[\displaystyle = - \frac{{\sqrt {{{\left[ {\sqrt 7 - \sqrt 2 } \right]}^2}} }}{{\sqrt {18} }}\] \[\displaystyle = - \frac{{\sqrt 7 - \sqrt 2 }}{{3\sqrt 2 }}\] \[\displaystyle = - \frac{{\sqrt {14} - 2}}{6}\]

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