Cho \(cos\alpha=\frac{\sqrt{2}}{3}\left(0< a< \frac{\pi}{2}\right)\). Tính giá trị \(cot\left(\alpha+\frac{3\pi}{2}\right)\)
Cho góc \(\alpha \) thỏa mãn \(\frac{\pi }{2} < \alpha < \pi ,\cos \alpha = - \frac{1}{{\sqrt 3 }}\). Tính giá trị của các biểu thức sau:
a) \(\sin \left( {\alpha + \frac{\pi }{6}} \right)\);
b) \(\cos \left( {\alpha + \frac{\pi }{6}} \right);\)
c) \(\sin \left( {\alpha - \frac{\pi }{3}} \right)\);
d) \(\cos \left( {\alpha - \frac{\pi }{6}} \right)\).
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) \(\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6} = \left( { - \frac{1}{{\sqrt 3 }}} \right).\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 6 }}{3}.\frac{1}{2} = \frac{{ - 3 + \sqrt 6 }}{6}\)
Cho \(\sin\alpha+\cos\alpha=\frac{\sqrt{6}}{2},a\in\left(0;\frac{\pi}{4}\right)\)
Tính giá trị biểu thức: \(P=\cos\left(\alpha+\frac{\pi}{4}\right)+\sqrt{2\left(1-\sin\alpha\cos\alpha+\sin\alpha-\cos\alpha\right)}\)
cho góc \(\alpha\) thỏa mãn \(\frac{\pi}{2}< \alpha< 2\pi\) và \(\cot\left(\alpha+\frac{\pi}{3}\right)=-\sqrt{3}\) tính giá trị biểu thức :
\(P=\sin\left(\alpha+\frac{\pi}{6}\right)+\cos\alpha\)
\(cot\left(a+\frac{\pi}{3}\right)=-\sqrt{3}\Leftrightarrow\frac{cos\left(a+\frac{\pi}{3}\right)}{sin\left(a+\frac{\pi}{3}\right)}=-\sqrt{3}\)
\(\Leftrightarrow cos\left(a+\frac{\pi}{3}\right)=-\sqrt{3}sin\left(a+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cosa.cos\frac{\pi}{3}-sina.sin\frac{\pi}{3}=-\sqrt{3}\left(sinacos\frac{\pi}{3}+cosa.sin\frac{\pi}{3}\right)\)
\(\Leftrightarrow\frac{1}{2}cosa-\frac{\sqrt{3}}{2}sina+\frac{\sqrt{3}}{2}sina+\frac{3}{2}cosa=0\)
\(\Leftrightarrow2cosa=0\Rightarrow cosa=0\Rightarrow a=\frac{3\pi}{2}\)
\(\Rightarrow P=sin\left(\frac{3\pi}{2}+\frac{\pi}{6}\right)+cos\frac{3\pi}{2}=-\frac{\sqrt{3}}{2}\)
Rút gọn các biểu thức sau:
a) \(\frac{1}{{\tan \alpha + 1}} + \frac{1}{{\cot \alpha + 1}}\)
b) \(\cos \left( {\frac{\pi }{2} - \alpha } \right) - \sin \left( {\pi + \alpha } \right)\)
c) \(\sin \left( {\alpha - \frac{\pi }{2}} \right) + \cos \left( { - \alpha + 6\pi } \right) - \tan \left( {\alpha + \pi } \right)\cot \left( {3\pi - \alpha } \right)\)
\(a,\dfrac{1}{tan\alpha+1}+\dfrac{1}{cot\alpha+1}\\ =\dfrac{cot\alpha+1+tan\alpha+1}{\left(tan\alpha+1\right)\left(cot\alpha+1\right)}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha\cdot cot\alpha+tan\alpha+cot\alpha+1}\\ =\dfrac{tan\alpha+cot\alpha+2}{tan\alpha+cot\alpha+2}\\ =1\)
\(b,cos\left(\dfrac{\pi}{2}-\alpha\right)-sin\left(\pi+\alpha\right)\\ =sin\alpha+sin\alpha\\ =2sin\alpha\)
\(c,sin\left(\alpha-\dfrac{\pi}{2}\right)+cos\left(-\alpha+6\pi\right)-tan\left(\alpha+\pi\right)cot\left(3\pi-\alpha\right)\\ =-sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\alpha\right)-tan\left(\alpha\right)cot\left(\pi-\alpha\right)\\ =-cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\alpha\right)\cdot cot\left(\alpha\right)\\ =1\)
1)Cho góc \(\alpha\) thõa mãn \(\frac{\pi}{2}< \alpha< 2\pi\) và \(tan\left(\alpha+\frac{\pi}{4}\right)=1\) . Tính P = \(cos\left(\pi-\frac{\pi}{6}\right)\) + \(sin\alpha\)
2)Cho góc \(\alpha\) thõa mãn \(\frac{\pi}{2}< \alpha< 2\pi\) và \(cot\left(\alpha+\frac{\pi}{3}\right)=-\sqrt{3}\) . Tính P = \(sin\left(\pi+\frac{\pi}{6}\right)\) + cos\(\alpha\)
Câu 1:
\(tan\left(a+\frac{\pi}{4}\right)=1\Rightarrow a+\frac{\pi}{4}=\frac{\pi}{4}+k\pi\Rightarrow a=k\pi\) (\(k\in Z\) )
Do \(\frac{\pi}{2}< a< 2\pi\Rightarrow\frac{\pi}{2}< k\pi< 2\pi\Rightarrow\frac{1}{2}< k< 2\Rightarrow k=1\Rightarrow a=\pi\)
\(\Rightarrow P=cos\left(\pi-\frac{\pi}{6}\right)+sin\pi=-\frac{\sqrt{3}}{2}\)
Câu 2:
\(cot\left(a+\frac{\pi}{3}\right)=-\sqrt{3}=cot\left(-\frac{\pi}{6}\right)\)
\(\Rightarrow a+\frac{\pi}{3}=-\frac{\pi}{6}+k\pi\Rightarrow a=-\frac{\pi}{2}+k\pi\) (\(k\in Z\))
\(\Rightarrow\frac{\pi}{2}< -\frac{\pi}{2}+k\pi< 2\pi\Rightarrow-\pi< k\pi< \frac{5\pi}{2}\)
\(\Rightarrow-1< k< \frac{5}{2}\Rightarrow k=\left\{0;1;2\right\}\Rightarrow a=\left\{-\frac{\pi}{2};\frac{\pi}{2};\frac{3\pi}{2}\right\}\) \(\Rightarrow cosa=0\)
\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)
Câu 1:
\(tan\left(a+\frac{\pi}{4}\right)=1\Leftrightarrow\frac{sin\left(a+\frac{\pi}{4}\right)}{cos\left(a+\frac{\pi}{4}\right)}=1\Leftrightarrow sin\left(a+\frac{\pi}{4}\right)=cos\left(a+\frac{\pi}{4}\right)\)
\(\Leftrightarrow sina.cos\frac{\pi}{4}+cosa.sin\frac{\pi}{4}=cosa.cos\frac{\pi}{4}-sina.sin\frac{\pi}{4}\)
\(\Leftrightarrow\frac{\sqrt{2}}{2}sina+\frac{\sqrt{2}}{2}cosa=\frac{\sqrt{2}}{2}cosa-\frac{\sqrt{2}}{2}sina\)
\(\Rightarrow\sqrt{2}sina=0\Rightarrow sina=0\)
\(\Rightarrow P=cos\left(\pi-\frac{\pi}{6}\right)+0=-cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}\)
Câu 2:
\(\frac{cos\left(a+\frac{\pi}{3}\right)}{sin\left(a+\frac{\pi}{3}\right)}=-\sqrt{3}\Leftrightarrow cos\left(a+\frac{\pi}{3}\right)=-\sqrt{3}sin\left(a+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(a+\frac{\pi}{3}\right)+\sqrt{3}sin\left(a+\frac{\pi}{3}\right)=0\)
\(\Leftrightarrow cosa.cos\frac{\pi}{3}-sina.sin\frac{\pi}{3}+\sqrt{3}sina.cos\frac{\pi}{3}+\sqrt{3}cosa.sin\frac{\pi}{3}=0\)
\(\Leftrightarrow\frac{1}{2}cosa-\frac{\sqrt{3}}{2}sina+\frac{\sqrt{3}}{2}sina+\frac{3}{2}cosa=0\)
\(\Leftrightarrow2cosa=0\Rightarrow cosa=0\)
\(\Rightarrow P=sin\left(\pi+\frac{\pi}{6}\right)+0=-sin\frac{\pi}{6}=-\frac{1}{2}\)
Cho \(cos\alpha = \frac{1}{3}\) và \( - \frac{\pi }{2} < \alpha < 0\). Tính
\(\begin{array}{l}a)\;sin\alpha \\b)\;sin2\alpha \\c)\;cos\left( {\alpha + \frac{\pi }{3}} \right)\end{array}\)
a, Ta có: \({\sin ^2}x + co{s^2}x = 1\)
\(\begin{array}{l} \Leftrightarrow {\sin ^2}\alpha + {\left( {\frac{1}{3}} \right)^2} = 1\\ \Leftrightarrow \sin \alpha = \pm \sqrt {1 - {{\left( {\frac{1}{3}} \right)}^2}} = \pm \frac{{2\sqrt 2 }}{3}\end{array}\)
Vì \( - \frac{\pi }{2} < \alpha < 0\) nên \(sin\alpha < 0 \Rightarrow \sin \alpha = - \frac{{2\sqrt 2 }}{3}\).
\(b)\;\,sin2\alpha = 2sin\alpha .cos\alpha = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} = - \frac{{4\sqrt 2 }}{9}\)
\(c)\;cos(\alpha + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6 + 1}}{6}\).
Tính \(\sin \left( {\alpha + \frac{\pi }{6}} \right),\cos \left( {\frac{\pi }{4} - \alpha } \right)\) biết \(\sin \alpha = - \frac{5}{{13}},\pi < \alpha < \frac{{3\pi }}{2}\)
\(\cos \alpha = - \sqrt {1 - {{\left( { - \frac{5}{{13}}} \right)}^2}} = - \frac{{12}}{{13}}\) (vì \(\pi < \alpha < \frac{{3\pi }}{2}\))
\(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)
\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha + \sin \frac{\pi }{4}sin\alpha = \frac{{ - 17\sqrt 2 }}{{26}}\)
a) Cho \(\sin\alpha=-\frac{3}{5}\left(\pi< \alpha< \frac{3\pi}{2}\right)\). Tính tan \(\alpha\)=?
b) Cho \(\alpha=\frac{\sqrt{3}}{3}\left(90^0< \alpha< 180^0\right)\). Tính cot \(\alpha\)=?
\(\pi< a< \frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\) \(\Rightarrow tana=\frac{sina}{cosa}=\frac{3}{4}\)
b/ \(sina=\frac{\sqrt{3}}{3}???cosa=\frac{\sqrt{3}}{3}???\)
a. \(\cos^2\alpha+\cos^2\left(\alpha-\frac{\pi}{3}\right)+^{ }\cos^2\left(\frac{2\pi}{3}-\alpha\right)=\frac{3}{2}\)