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Câu hỏi: Tính
Phương pháp giải:
+) Với \(0 < \alpha < \dfrac{\pi }{2}\) ta có: \(\sin \alpha >0, \cos \alpha >0.\)
+) Với \( \dfrac{\pi }{2} < \alpha < \pi \) ta có: \(\sin \alpha >0, \cos \alpha < 0.\)
+) \(\sin^2 \alpha +\cos^2 \alpha =1. \)
+) \({\tan ^2}x + 1 = \dfrac{1}{{{{\cos }^2}x}}.\)
+) \({\cot ^2}x + 1 = \dfrac{1}{{{{\sin }^2}x}}.\)
+) \(\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\)
+) \(\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .\)
+) \( \cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta .\)
+) \(\tan\left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}.\)
+) \(\tan\left( {\alpha - \beta } \right) = \dfrac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}.\)
Lời giải chi tiết:
Ta có:
\(\begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \Rightarrow {\cos ^2}\alpha = 1 - {\sin ^2}\alpha \\ = 1 - {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} = \dfrac{2}{3}\end{array}\)
Mà \(0 < \alpha < \dfrac{\pi }{2} \Rightarrow \cos \alpha > 0\)
\(\Rightarrow \cos \alpha = \sqrt {\dfrac{2}{3}} = \dfrac{{\sqrt 6 }}{3}\)
\(\Rightarrow \cos(α + \dfrac{\pi}{3}) = \cosα\cos \dfrac{\pi }{3} - \sinα\sin \dfrac{\pi}{3}\)
\(= \dfrac{\sqrt{6}}{3}.\dfrac{1}{2}-\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{6}-3}{6}\)
Lời giải chi tiết:
\(\begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \Rightarrow {\sin ^2}\alpha = 1 - {\cos ^2}\alpha \\ = 1 - {\left( { - \dfrac{1}{3}} \right)^2} = \dfrac{8}{9}\end{array}\)
Mà \(\dfrac{\pi }{2} < \alpha < \pi \Rightarrow \sin \alpha > 0\)
\(\Rightarrow \sin \alpha = \sqrt {\dfrac{8}{9}} = \dfrac{{2\sqrt 2 }}{3}\)
\(\Rightarrow \tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}\) \(= \dfrac{{2\sqrt 2 }}{3}:\left( { - \dfrac{1}{3}} \right) = - 2\sqrt 2 \)
\(\tan(α - \dfrac{\pi}{4}) \) \(= \dfrac{\tan\alpha -\tan\dfrac{\pi}{4}}{1+\tan\alpha \tan\dfrac{\pi}{4}}\) \(=\dfrac{-1-2\sqrt{2}}{1-2\sqrt{2}}\) \(=\dfrac{2\sqrt{2}+1}{2\sqrt{2}-1}\) \( = \dfrac{{{{\left( {2\sqrt 2 + 1} \right)}^2}}}{{8 - 1}}\) \(= \dfrac{{9 + 4\sqrt 2 }}{7}.\)
Lời giải chi tiết:
\(\begin{array}{l}{\sin ^2}a + {\cos ^2}a = 1\\ \Rightarrow {\cos ^2}a = 1 - {\sin ^2}a\\ = 1 - {\left( {\dfrac{4}{5}} \right)^2} = \dfrac{9}{{25}}\end{array}\)
Mà \({0^0} < a < {90^0} \Rightarrow \cos a > 0\) \(\Rightarrow \cos a = \sqrt {\dfrac{9}{{25}}} = \dfrac{3}{5}\).
\(\begin{array}{l}{\sin ^2}b + {\cos ^2}b = 1\\ \Rightarrow {\cos ^2}b = 1 - {\sin ^2}b\\ = 1 - {\left( {\dfrac{2}{3}} \right)^2} = \dfrac{5}{9}\end{array}\)
Mà \({90^0} < b < {180^0} \Rightarrow \cos b < 0\) \(\Rightarrow \cos b = - \sqrt {\dfrac{5}{9}} = - \dfrac{{\sqrt 5 }}{3}\).
\(\cos(a + b) = \cos a\cos b - \sin a\sin b\)
\(=\dfrac{3}{5}\left ( -\dfrac{\sqrt{5}}{3} \right)-\dfrac{4}{5}.\dfrac{2}{3}=-\dfrac{3\sqrt{5}+8}{15}\)
Câu a
\(\cos(α + \dfrac{\pi}{3}),\) biết \(\sinα = \dfrac{1}{\sqrt{3}}\) và \(0 < α < \dfrac{\pi }{2}.\)Phương pháp giải:
+) Với \(0 < \alpha < \dfrac{\pi }{2}\) ta có: \(\sin \alpha >0, \cos \alpha >0.\)
+) Với \( \dfrac{\pi }{2} < \alpha < \pi \) ta có: \(\sin \alpha >0, \cos \alpha < 0.\)
+) \(\sin^2 \alpha +\cos^2 \alpha =1. \)
+) \({\tan ^2}x + 1 = \dfrac{1}{{{{\cos }^2}x}}.\)
+) \({\cot ^2}x + 1 = \dfrac{1}{{{{\sin }^2}x}}.\)
+) \(\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .\)
+) \(\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .\)
+) \( \cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta .\)
+) \(\tan\left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}.\)
+) \(\tan\left( {\alpha - \beta } \right) = \dfrac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}.\)
Lời giải chi tiết:
Ta có:
\(\begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \Rightarrow {\cos ^2}\alpha = 1 - {\sin ^2}\alpha \\ = 1 - {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2} = \dfrac{2}{3}\end{array}\)
Mà \(0 < \alpha < \dfrac{\pi }{2} \Rightarrow \cos \alpha > 0\)
\(\Rightarrow \cos \alpha = \sqrt {\dfrac{2}{3}} = \dfrac{{\sqrt 6 }}{3}\)
\(\Rightarrow \cos(α + \dfrac{\pi}{3}) = \cosα\cos \dfrac{\pi }{3} - \sinα\sin \dfrac{\pi}{3}\)
\(= \dfrac{\sqrt{6}}{3}.\dfrac{1}{2}-\dfrac{1}{\sqrt{3}}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{6}-3}{6}\)
Câu b
\(\tan(α - \dfrac{\pi }{4}),\) biết \(\cosα = -\dfrac{1}{3}\) và \(\dfrac{\pi }{2} < α < π.\)Lời giải chi tiết:
\(\begin{array}{l}{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\ \Rightarrow {\sin ^2}\alpha = 1 - {\cos ^2}\alpha \\ = 1 - {\left( { - \dfrac{1}{3}} \right)^2} = \dfrac{8}{9}\end{array}\)
Mà \(\dfrac{\pi }{2} < \alpha < \pi \Rightarrow \sin \alpha > 0\)
\(\Rightarrow \sin \alpha = \sqrt {\dfrac{8}{9}} = \dfrac{{2\sqrt 2 }}{3}\)
\(\Rightarrow \tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}\) \(= \dfrac{{2\sqrt 2 }}{3}:\left( { - \dfrac{1}{3}} \right) = - 2\sqrt 2 \)
\(\tan(α - \dfrac{\pi}{4}) \) \(= \dfrac{\tan\alpha -\tan\dfrac{\pi}{4}}{1+\tan\alpha \tan\dfrac{\pi}{4}}\) \(=\dfrac{-1-2\sqrt{2}}{1-2\sqrt{2}}\) \(=\dfrac{2\sqrt{2}+1}{2\sqrt{2}-1}\) \( = \dfrac{{{{\left( {2\sqrt 2 + 1} \right)}^2}}}{{8 - 1}}\) \(= \dfrac{{9 + 4\sqrt 2 }}{7}.\)
Câu c
\(\cos(a + b), \sin(a - b)\) biết \(\sin a = \frac{4}{5}\) \(0^0< a < 90^0,\) và \(\sin b = \frac{2}{3},\) \(90^0< b < 180^0.\) Lời giải chi tiết:
\(\begin{array}{l}{\sin ^2}a + {\cos ^2}a = 1\\ \Rightarrow {\cos ^2}a = 1 - {\sin ^2}a\\ = 1 - {\left( {\dfrac{4}{5}} \right)^2} = \dfrac{9}{{25}}\end{array}\)
Mà \({0^0} < a < {90^0} \Rightarrow \cos a > 0\) \(\Rightarrow \cos a = \sqrt {\dfrac{9}{{25}}} = \dfrac{3}{5}\).
\(\begin{array}{l}{\sin ^2}b + {\cos ^2}b = 1\\ \Rightarrow {\cos ^2}b = 1 - {\sin ^2}b\\ = 1 - {\left( {\dfrac{2}{3}} \right)^2} = \dfrac{5}{9}\end{array}\)
Mà \({90^0} < b < {180^0} \Rightarrow \cos b < 0\) \(\Rightarrow \cos b = - \sqrt {\dfrac{5}{9}} = - \dfrac{{\sqrt 5 }}{3}\).
\(\cos(a + b) = \cos a\cos b - \sin a\sin b\)
\(=\dfrac{3}{5}\left ( -\dfrac{\sqrt{5}}{3} \right)-\dfrac{4}{5}.\dfrac{2}{3}=-\dfrac{3\sqrt{5}+8}{15}\)
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