Bài 6 trang 154 SGK Đại số 10

Đề bài

Cho $\displaystyle \sin 2a =  - {5 \over 9}$ và $\displaystyle {\pi  \over 2} < a < π$.

Tính $\displaystyle \sin a$ và $\displaystyle \cos a.$

Lời giải chi tiết

Với $\displaystyle {\pi  \over 2}< a < π\Rightarrow \sin a > 0, \cos a < 0.$

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

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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