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Câu hỏi: Cho \(\displaystyle \sin 2a = - {5 \over 9}\) và \(\displaystyle {\pi \over 2} < a < π\).
Tính \(\displaystyle \sin a\) và \(\displaystyle \cos a.\)
Tính \(\displaystyle \sin a\) và \(\displaystyle \cos a.\)
Phương pháp giải
+) Với \(\dfrac{{\pi }}{2} < a < \pi\) ta có \(\sin a > 0, \cos a < 0.\)
+) \(\sin^2 \alpha +\cos^2 \alpha =1. \)
+) \(\sin 2a = 2\sin a.\cos a.\)
+) \(\cos 2a = {\cos ^2}a - {\sin ^2}a \) \(= 2{\cos ^2}a - 1\)\(= 1 - 2{\sin ^2}a.\)
+) \(\sin^2 a = \dfrac{{1 - \cos 2a}}{2}.\)
+) \(\cos^2 a = \dfrac{{1 + \cos 2a}}{2}.\)
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
\(\displaystyle {\sin ^2}a = \frac{{1 - \cos 2a}}{2}\)
\(\displaystyle \eqalign{
& \Rightarrow \sin a = \sqrt {{{1 - \cos 2a} \over 2}} = \sqrt {{{1 + {{2\sqrt {14} } \over 9}} \over 2}} \cr&= {{\sqrt {9 + 2\sqrt {14} } } \over {3\sqrt 2 }} = {{\sqrt {{{\left({\sqrt 7 + \sqrt 2 } \right)}^2}} } \over {3\sqrt 2 }}\cr& = {{\sqrt 7 + \sqrt 2 } \over {3\sqrt 2 }} = {{\sqrt {14} + 2} \over 6} \cr } \)
\(\displaystyle {\cos ^2}a = \frac{{1 + \cos 2a}}{2}\)
\(\displaystyle \Rightarrow \cos a = - \sqrt {{{1 + \cos 2a} \over 2}} \) \(\displaystyle = - \sqrt {\frac{{1 - \frac{{2\sqrt {14} }}{9}}}{2}} \) \(\displaystyle = - \sqrt {\frac{{9 - 2\sqrt {14} }}{{18}}} \) \(\displaystyle = - \frac{{\sqrt {{{\left( {\sqrt 7 - \sqrt 2 } \right)}^2}} }}{{\sqrt {18} }}\) \(\displaystyle = - \frac{{\sqrt 7 - \sqrt 2 }}{{3\sqrt 2 }} \) \(\displaystyle = - \frac{{\sqrt {14} - 2}}{6}\)
Cách khác:
Ta có:
$\frac{\pi}{2}<a<\pi \Rightarrow \sin a>0$ và $\cos a<0 .$
$+(\sin a-\cos a)^{2}$
$=\sin ^{2} a+\cos ^{2} a-2 \cdot \sin a \cdot \cos a$
$=1-\sin 2 a=1-\frac{-5}{9}=\frac{14}{9}$
Mà $\sin a>0 ; \cos a<0$ nên
$\sin a-\cos a>0 \Rightarrow \sin a-\cos a=\frac{\sqrt{14}}{3} (1)$
$+(\sin a+\cos a)^{2}$
$=\sin ^{2} a+\cos ^{2} a+2 \cdot \sin a \cdot \cos a$
$=1+\sin 2 a=1+\frac{-5}{9}=\frac{4}{9}$
$\Rightarrow \sin a+\cos a=\frac{2}{3}$
hoặc $\sin a+\cos a=\frac{-2}{3}$.
TH1: $\sin a+\cos a=\frac{2}{3}$.
Kết hợp với (1) ta được hệ phương trình:
$\left\{\begin{array}{l}\sin a+\cos a=\frac{2}{3} \\ \sin a-\cos a=\frac{\sqrt{14}}{3}\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin a=\frac{2+\sqrt{14}}{6} \\ \cos a=\frac{2-\sqrt{14}}{6}\end{array}\right.\right.$
$\mathrm{TH} 2: \sin a+\cos a=\frac{-2}{3}$
Kết hợp với (1) ta được hệ phương trình:
$\left\{\begin{array}{l}\sin a+\cos a=\frac{-2}{3} \\ \sin a-\cos a=\frac{\sqrt{14}}{3}\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin a=\frac{-2+\sqrt{14}}{6} \\ \cos a=\frac{-2-\sqrt{14}}{6}\end{array}\right.\right.$
+) Với \(\dfrac{{\pi }}{2} < a < \pi\) ta có \(\sin a > 0, \cos a < 0.\)
+) \(\sin^2 \alpha +\cos^2 \alpha =1. \)
+) \(\sin 2a = 2\sin a.\cos a.\)
+) \(\cos 2a = {\cos ^2}a - {\sin ^2}a \) \(= 2{\cos ^2}a - 1\)\(= 1 - 2{\sin ^2}a.\)
+) \(\sin^2 a = \dfrac{{1 - \cos 2a}}{2}.\)
+) \(\cos^2 a = \dfrac{{1 + \cos 2a}}{2}.\)
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
\(\displaystyle {\sin ^2}a = \frac{{1 - \cos 2a}}{2}\)
\(\displaystyle \eqalign{
& \Rightarrow \sin a = \sqrt {{{1 - \cos 2a} \over 2}} = \sqrt {{{1 + {{2\sqrt {14} } \over 9}} \over 2}} \cr&= {{\sqrt {9 + 2\sqrt {14} } } \over {3\sqrt 2 }} = {{\sqrt {{{\left({\sqrt 7 + \sqrt 2 } \right)}^2}} } \over {3\sqrt 2 }}\cr& = {{\sqrt 7 + \sqrt 2 } \over {3\sqrt 2 }} = {{\sqrt {14} + 2} \over 6} \cr } \)
\(\displaystyle {\cos ^2}a = \frac{{1 + \cos 2a}}{2}\)
\(\displaystyle \Rightarrow \cos a = - \sqrt {{{1 + \cos 2a} \over 2}} \) \(\displaystyle = - \sqrt {\frac{{1 - \frac{{2\sqrt {14} }}{9}}}{2}} \) \(\displaystyle = - \sqrt {\frac{{9 - 2\sqrt {14} }}{{18}}} \) \(\displaystyle = - \frac{{\sqrt {{{\left( {\sqrt 7 - \sqrt 2 } \right)}^2}} }}{{\sqrt {18} }}\) \(\displaystyle = - \frac{{\sqrt 7 - \sqrt 2 }}{{3\sqrt 2 }} \) \(\displaystyle = - \frac{{\sqrt {14} - 2}}{6}\)
Cách khác:
Ta có:
$\frac{\pi}{2}<a<\pi \Rightarrow \sin a>0$ và $\cos a<0 .$
$+(\sin a-\cos a)^{2}$
$=\sin ^{2} a+\cos ^{2} a-2 \cdot \sin a \cdot \cos a$
$=1-\sin 2 a=1-\frac{-5}{9}=\frac{14}{9}$
Mà $\sin a>0 ; \cos a<0$ nên
$\sin a-\cos a>0 \Rightarrow \sin a-\cos a=\frac{\sqrt{14}}{3} (1)$
$+(\sin a+\cos a)^{2}$
$=\sin ^{2} a+\cos ^{2} a+2 \cdot \sin a \cdot \cos a$
$=1+\sin 2 a=1+\frac{-5}{9}=\frac{4}{9}$
$\Rightarrow \sin a+\cos a=\frac{2}{3}$
hoặc $\sin a+\cos a=\frac{-2}{3}$.
TH1: $\sin a+\cos a=\frac{2}{3}$.
Kết hợp với (1) ta được hệ phương trình:
$\left\{\begin{array}{l}\sin a+\cos a=\frac{2}{3} \\ \sin a-\cos a=\frac{\sqrt{14}}{3}\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin a=\frac{2+\sqrt{14}}{6} \\ \cos a=\frac{2-\sqrt{14}}{6}\end{array}\right.\right.$
$\mathrm{TH} 2: \sin a+\cos a=\frac{-2}{3}$
Kết hợp với (1) ta được hệ phương trình:
$\left\{\begin{array}{l}\sin a+\cos a=\frac{-2}{3} \\ \sin a-\cos a=\frac{\sqrt{14}}{3}\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin a=\frac{-2+\sqrt{14}}{6} \\ \cos a=\frac{-2-\sqrt{14}}{6}\end{array}\right.\right.$