Cho tan2α = 2 và π < α < \(\frac{3\pi}{2}\). Biết giá trị của biểu thức M= \(\frac{cos(\alpha+\frac{\pi}{3})+cos(\alpha-\frac{\pi}{3})}{tan(\frac{\pi}{2}-\alpha)+tan(\frac{\pi+\alpha}{2}}=\frac{a}{\sqrt{b}}\) với a, b là các số nguyên. Khi đó, giá trị của biểu thức T = 2a + b là ?
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a) Vì \(0<\alpha <\frac{\pi }{2} \) nên \(\sin \alpha > 0\). Mặt khác, từ \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) suy ra
\(\sin \alpha = \sqrt {1 - {{\cos }^2}a} = \sqrt {1 - \frac{1}{{25}}} = \frac{{2\sqrt 6 }}{5}\)
Do đó, \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{{2\sqrt 6 }}{5}}}{{\frac{1}{5}}} = 2\sqrt 6 \) và \(\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{\frac{1}{5}}}{{\frac{{2\sqrt 6 }}{5}}} = \frac{{\sqrt 6 }}{{12}}\)
b) Vì \(\frac{\pi }{2} < \alpha < \pi\) nên \(\cos \alpha < 0\). Mặt khác, từ \({\sin ^2}\alpha + {\cos ^2}\alpha = 1\) suy ra
\(\cos \alpha = \sqrt {1 - {{\sin }^2}a} = \sqrt {1 - \frac{4}{9}} = -\frac{{\sqrt 5 }}{3}\)
Do đó, \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{\frac{2}{3}}}{{-\frac{{\sqrt 5 }}{3}}} = -\frac{{2\sqrt 5 }}{5}\) và \(\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{{-\frac{{\sqrt 5 }}{3}}}{{\frac{2}{3}}} = -\frac{{\sqrt 5 }}{2}\)
c) Ta có: \(\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{{\sqrt 5 }}\)
Ta có: \({\tan ^2}\alpha + 1 = \frac{1}{{{{\cos }^2}\alpha }} \Rightarrow {\cos ^2}\alpha = \frac{1}{{{{\tan }^2}\alpha + 1}} = \frac{1}{6} \Rightarrow \cos \alpha = \pm \frac{1}{{\sqrt 6 }}\)
Vì \(\pi < \alpha < \frac{{3\pi }}{2} \Rightarrow \sin \alpha < 0\;\) và \(\,\,\cos \alpha < 0 \Rightarrow \cos \alpha = -\frac{1}{{\sqrt 6 }}\)
Ta có: \(\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} \Rightarrow \sin \alpha = \tan \alpha .\cos \alpha = \sqrt 5 .(-\frac{1}{{\sqrt 6 }}) = -\sqrt {\frac{5}{6}} \)
d) Vì \(\cot \alpha = - \frac{1}{{\sqrt 2 }}\;\,\) nên \(\,\,\tan \alpha = \frac{1}{{\cot \alpha }} = - \sqrt 2 \)
Ta có: \({\cot ^2}\alpha + 1 = \frac{1}{{{{\sin }^2}\alpha }} \Rightarrow {\sin ^2}\alpha = \frac{1}{{{{\cot }^2}\alpha + 1}} = \frac{2}{3} \Rightarrow \sin \alpha = \pm \sqrt {\frac{2}{3}} \)
Vì \(\frac{{3\pi }}{2} < \alpha < 2\pi \Rightarrow \sin \alpha < 0 \Rightarrow \sin \alpha = - \sqrt {\frac{2}{3}} \)
Ta có: \(\cot \alpha = \frac{{\cos \alpha }}{{\sin \alpha }} \Rightarrow \cos \alpha = \cot \alpha .\sin \alpha = \left( { - \frac{1}{{\sqrt 2 }}} \right).\left( { - \sqrt {\frac{2}{3}} } \right) = \frac{{\sqrt 3 }}{3}\)
Ta có:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha \sin \frac{\pi }{6} = \frac{{\sqrt 6 }}{3}.\frac{{\sqrt 3 }}{2} + \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{1}{2} = \frac{{ - \sqrt 3 + 3\sqrt 2 }}{6}\)
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right) = \cos \alpha .\cos \frac{\pi }{6} - \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 6 }}{3}.\frac{1}{2} = - \frac{{3 + \sqrt 6 }}{6}\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right) = \sin \alpha \cos \frac{\pi }{3} - \cos \alpha \sin \frac{\pi }{3} = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} = \frac{{3 + \sqrt 6 }}{6}\)
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
vậy thì kết quả là
\(\sin2\alpha=-0.96\)
\(\)còn \(\cos\left(\alpha+\frac{\pi}{6}\right)\) thì đúng vì -(-0.8) mà sorry thiếu ngủ hôm qua -_-
Ta có:
\(\begin{array}{l}{\tan ^2}\alpha + 1 = \frac{1}{{{{\cos }^2}\alpha }}\\ \Rightarrow {\left( {\frac{2}{3}} \right)^2} + 1 = \frac{1}{{{{\cos }^2}\alpha }}\\ \Rightarrow \frac{1}{{{{\cos }^2}\alpha }} = \frac{{13}}{9}\\ \Rightarrow \cos \alpha = \pm \frac{{3\sqrt {13} }}{{13}}\end{array}\)
Do \(\pi < \alpha < \frac{{3\pi }}{2} \Rightarrow \cos \alpha = - \frac{{3\sqrt {13} }}{{13}}\)
Ta có: \(\begin{array}{l}\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} \Rightarrow \frac{2}{3} = \sin \alpha :\left( { - \frac{{3\sqrt {13} }}{{13}}} \right)\\ \Rightarrow \sin \alpha = - \frac{{2\sqrt {13} }}{{13}}\end{array}\)
a/ \(\frac{\pi}{6}< x< \frac{\pi}{3}\Rightarrow cosx>0\)
\(cos^2x=\frac{1}{1+tan^2x}=\frac{1}{10}\)
\(cotx=\frac{1}{tanx}=\frac{1}{3}\)
Thay số và bấm máy
b/ \(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\tana< 0\end{matrix}\right.\)
\(sina=\sqrt{1-cos^2a}=\frac{3}{5}\)
\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)
\(A=\frac{6sina.cosa-\frac{2tana}{1-tan^2a}}{cosa-\left(2cos^2a-1\right)}\)
Thay số và bấm máy
c/ \(\frac{3\pi}{2}< x< 2\pi\Rightarrow\left\{{}\begin{matrix}cosx>0\\sinx< 0\end{matrix}\right.\)
\(cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{1}{\sqrt{5}}\)
\(sinx=cosx.tanx=-\frac{2}{\sqrt{5}}\)
\(B=\frac{cos^2x+2sinx.cosx}{\frac{2tanx}{1-tan^2x}-\left(2cos^2x-1\right)}\)
Thay số
a) Ta có \({\cos ^2}\alpha + {\sin ^2}\alpha \,\,\, = \,1\)
mà \(\sin \alpha = \frac{{\sqrt {15} }}{4}\) nên \({\cos ^2}\alpha + {\left( {\frac{{\sqrt {15} }}{4}} \right)^2}\,\,\, = \,1 \Rightarrow {\cos ^2}\alpha = \frac{1}{{16}}\)
Lại có \(\frac{\pi }{2} < \alpha < \pi \) nên \(\cos \alpha < 0 \Rightarrow \cos \alpha = - \frac{1}{4}\)
Khi đó \(\tan \alpha = \frac{{\sin \alpha }}{{co{\mathop{\rm s}\nolimits} \alpha }} = - \sqrt {15} ;\cot \alpha = \frac{1}{{\tan \alpha }} = - \frac{1}{{\sqrt {15} }}\)
b)
Ta có \({\cos ^2}\alpha + {\sin ^2}\alpha \,\,\, = \,1\)
mà \(\cos \alpha = - \frac{2}{3}\) nên \({\sin ^2}\alpha + {\left( {\frac{{ - 2}}{3}} \right)^2}\,\,\, = \,1 \Rightarrow {\sin ^2}\alpha = \frac{5}{9}\)
Lại có \( - \pi < \alpha < 0\) nên \(\sin \alpha < 0 \Rightarrow \sin \alpha = - \frac{{\sqrt 5 }}{3}\)
Khi đó \(\tan \alpha = \frac{{\sin \alpha }}{{co{\mathop{\rm s}\nolimits} \alpha }} = \frac{{\sqrt 5 }}{2};\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{2}{{\sqrt 5 }}\)
c)
Ta có \(\tan \alpha = 3\) nên
\(\cot \alpha = \frac{1}{{\tan \alpha }} = \frac{1}{3}\)
\(\frac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha \,\,\, = \,1 + {3^2} = 10\,\, \Rightarrow {\cos ^2}\alpha = \frac{1}{{10}}\)
Mà \({\cos ^2}\alpha + {\sin ^2}\alpha \,\,\, = \,1 \Rightarrow {\sin ^2}\alpha = \frac{9}{{10}}\)
Với \( - \pi < \alpha < 0\) thì \(\sin \alpha < 0 \Rightarrow \sin \alpha = - \sqrt {\frac{9}{{10}}} \)
Với \( - \pi < \alpha < - \frac{\pi }{2}\) thì \(\cos \alpha < 0 \Rightarrow \cos \alpha = - \sqrt {\frac{1}{{10}}} \)
và \( - \frac{\pi }{2} \le \alpha < 0\) thì \(\cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {\frac{1}{{10}}} \)
d)
Ta có \(\cot \alpha = - 2\) nên
\(\tan \alpha = \frac{1}{{\cot \alpha }} = - \frac{1}{2}\)
\(\frac{1}{{{{\sin }^2}\alpha }} = 1 + co{{\mathop{\rm t}\nolimits} ^2}\alpha \,\,\, = \,1 + {( - 2)^2} = 5\,\, \Rightarrow {\sin ^2}\alpha = \frac{1}{5}\)
Mà \({\cos ^2}\alpha + {\sin ^2}\alpha \,\,\, = \,1 \Rightarrow {\cos ^2}\alpha = \frac{4}{5}\)
Với \(0 < \alpha < \pi \) thì \(\sin \alpha > 0 \Rightarrow \sin \alpha = \sqrt {\frac{1}{5}} \)
Với \(0 < \alpha < \frac{\pi }{2}\) thì \(\cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {\frac{4}{5}} \)
và \(\frac{\pi }{2} \le \alpha < \pi \) thì \(\cos \alpha < 0 \Rightarrow \cos \alpha = - \sqrt {\frac{4}{5}} \)
a, \(sin\alpha=\frac{1}{5},\frac{\pi}{2}< \alpha< \pi\)
+) \(sin^2\alpha+cos^2\alpha=1\)
\(\Leftrightarrow\left(\frac{1}{5}\right)^2+cos^2\alpha=1\Leftrightarrow cos^2\alpha=\frac{24}{25}\Leftrightarrow cos\alpha=\pm\frac{2\sqrt{6}}{5}\)
mà \(\frac{\pi}{2}< \alpha< \pi\Rightarrow cos\alpha=-\frac{2\sqrt{6}}{5}\)
+) \(tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{\frac{1}{5}}{-\frac{2\sqrt{6}}{5}}=-\frac{\sqrt{6}}{12}\)
+) \(cot\alpha=\frac{cos\alpha}{sin\alpha}=\frac{-\frac{2\sqrt{6}}{5}}{\frac{1}{5}}=-2\sqrt{6}\)
a/ \(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{2\sqrt{6}}{5}\)
\(tanx=\frac{sinx}{cosx}=-\frac{\sqrt{6}}{12}\) ; \(cotx=\frac{1}{tanx}=-2\sqrt{6}\)
b/ \(\frac{3\pi}{2}< a< 2\pi\Rightarrow cosa>0\)
\(\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}=\frac{5\sqrt{26}}{26}\)
\(sina=tana.cosa=-\frac{\sqrt{26}}{26}\)
c/ \(0< a< \frac{\pi}{2}\Rightarrow sina;cosa>0\)
\(\left\{{}\begin{matrix}cos^2a+sin^2a=1\\2sina.cosa=\frac{2}{3}\end{matrix}\right.\)
\(\Rightarrow sina+cosa=\frac{\sqrt{15}}{3}\Rightarrow cosa=\frac{\sqrt{15}}{3}-sina\)
\(\Rightarrow sina\left(\frac{\sqrt{15}}{3}-sina\right)=\frac{1}{3}\Rightarrow sin^2a-\frac{\sqrt{15}}{3}sina+\frac{1}{3}=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=\frac{\sqrt{15}+\sqrt{3}}{6}\Rightarrow cosa=\frac{\sqrt{15}-\sqrt{3}}{6}\\sina=\frac{\sqrt{15}-\sqrt{3}}{6}\Rightarrow cosa=\frac{\sqrt{15}+\sqrt{3}}{6}\end{matrix}\right.\) \(\Rightarrow tana=\frac{sina}{cosa}=...\)
d/ \(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\cosa< 0\end{matrix}\right.\)
\(cosa=\sqrt{2}-sina\) \(\Rightarrow sin^2a+\left(\sqrt{2}-sina\right)^2=1\)
\(\Leftrightarrow2sin^2a-2\sqrt{2}sina+1=0\Rightarrow sina=\frac{\sqrt{2}}{2}\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{\sqrt{2}}{2}\)
\(tana=\frac{sina}{cosa}=-1\)
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
\(\pi< a< \frac{3\pi}{2}\Rightarrow2\pi< 2a< 3\pi\Rightarrow sin2a>0\)
\(cot2a=\frac{1}{2}\Rightarrow sin2a=\frac{1}{\sqrt{1+cot^22a}}=\frac{2\sqrt{5}}{5}\)
\(cos\left(a+\frac{\pi}{3}\right)+cos\left(a-\frac{\pi}{3}\right)=2cosa.cos\frac{\pi}{3}=cosa\)
\(tan\left(\frac{\pi}{2}-a\right)+tan\left(\frac{\pi}{2}+\frac{a}{2}\right)=\frac{-sin\frac{a}{2}}{cos\left(\frac{\pi}{2}-a\right).cos\left(\frac{\pi}{2}+\frac{a}{2}\right)}=\frac{sin\frac{a}{2}}{sina.sin\frac{a}{2}}=\frac{1}{sina}\)
\(\Rightarrow M=sina.cosa=\frac{1}{2}sin2a=\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow2a+b=7\)