a Cho \(\sin\alpha=\frac{3}{5}\) , \(0< \alpha< \frac{\pi}{2}\). Tính \(\sin\left(\alpha+\frac{\pi}{6}\right)\), \(\sin2\alpha\)
b Cho \(\sin\alpha=-\frac{4}{5}\),\(\frac{\pi}{2}< \alpha< \pi\). Tính \(\cos\left(\alpha-\frac{\pi}{3}\right)\),\(\cos2\alpha\)
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 \(\pi < \alpha < \frac{{3\pi }}{2}\))
\(\sin \left( {\alpha + \frac{\pi }{6}} \right) = \sin \alpha \cos \frac{\pi }{6} + \cos \alpha sin\frac{\pi }{6} = \frac{{ - 12 + 5\sqrt 3 }}{{26}}\)
\(\cos \left( {\frac{\pi }{4} - \alpha } \right) = \cos \frac{\pi }{4}\cos \alpha + \sin \frac{\pi }{4}sin\alpha = \frac{{ - 17\sqrt 2 }}{{26}}\)
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 \(sin\alpha < 0 \Rightarrow \sin \alpha = - \frac{{2\sqrt 2 }}{3}\).
\(b)\;\,sin2\alpha = 2sin\alpha .cos\alpha = 2.\left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{1}{3} = - \frac{{4\sqrt 2 }}{9}\)
\(c)\;cos(\alpha + \frac{\pi }{3}) = cos\alpha .cos\frac{\pi }{3} - sin\alpha .sin\frac{\pi }{3}\)\( = \frac{1}{3}.\frac{1}{2} - \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} = \frac{{2\sqrt 6 + 1}}{6}\).
Cho cos \(\alpha\)=\(-\frac{4}{5}\) và \(-\pi< \alpha< \frac{-3}{2}\pi\). Tính \(\sin2\alpha;\)\(\cos2a;\sin\left(\frac{5\pi}{2}-\alpha\right);\tan\left(\alpha+\frac{\pi}{4}\right);\cos\frac{\alpha}{2}\)
--.-- \(-\pi>-\frac{3}{2}\pi\) mà
Chắc nhầm đề rồi, phải là \(-\pi>a>-\frac{3}{2}\pi\)mới đúng chứ
\(-\pi>a>-\frac{3}{2}\pi\Leftrightarrow\pi>a>\frac{1}{2}\pi\)
\(\cos a=-\frac{4}{5}\Rightarrow\sin a=\frac{3}{5}\)
\(\sin2a=2\sin a.\cos a=2.\frac{3}{5}.\frac{-4}{5}=-\frac{24}{25}\)
\(\cos2a=2\cos^2a-1=\frac{7}{25}\)
\(\sin\left(\frac{5\pi}{2}-a\right)=\sin\left(\frac{\pi}{2}-a\right)=\cos a=-\frac{4}{5}\)
\(\sin\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{3}{5}-\frac{4}{5}.\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{10}\)
\(\cos\left(a+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}.\frac{-4}{5}-\frac{\sqrt{2}}{2}.\frac{3}{5}=-\frac{7\sqrt{2}}{10}\)
\(\Rightarrow\tan\left(a+\frac{\pi}{4}\right)=\frac{1}{7}\)
\(\cos^2\left(\frac{a}{2}\right)=\frac{1+\cos a}{2}=\frac{1}{10}\Leftrightarrow\left|\cos\frac{a}{2}\right|=\frac{\sqrt{10}}{10}\)
Mà \(\frac{\pi}{2}>\frac{a}{2}>\frac{\pi}{4}\)
\(\Rightarrow\cos\frac{a}{2}=\frac{\sqrt{10}}{10}\)
1) Cho sinα = \(\frac{3}{5}\) và \(\frac{\pi}{2}\)<α<π
a) cos α, tanα, cotα
b) sin(α - \(\frac{\pi}{3}\)) ; cos2α
2) cho cosα = 0,6 và \(\frac{3\pi}{2}\)<α<2π
a) sinα, tanα, cotα
b) sin2α ; cos(α + \(\frac{\pi}{6}\))
Cho \(\sin \alpha = \frac{{12}}{{13}}\) và \(\cos \alpha = - \frac{5}{{13}}\). Tính \(\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi + \alpha } \right)\)
Ta có:
\(\begin{array}{l}\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi + \alpha } \right) = \sin \left( { -\frac{{16\pi }}{2} +\frac{{\pi }}{2} + \alpha } \right) - \cos \left( {12\pi + \pi + \alpha } \right) = \sin \left( {-8\pi + \frac{\pi }{2} - \alpha } \right) - \cos \left( { \pi + \alpha } \right) \\ = \sin \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( \alpha \right) = \cos \left( \alpha \right) + \cos \left( \alpha \right) = 2\cos \left( \alpha \right) = 2.\left( { - \frac{5}{{13}}} \right) = \frac{{ - 10}}{{13}}\end{array}\)
Cho \(\alpha\in\left(\frac{\Pi}{2};\Pi\right)\) và \(sin\alpha=\frac{3}{5}\). Tính \(A=\frac{sin\left(\frac{7\Pi}{2}-\alpha\right)}{sin\left(\frac{\Pi}{4}+\alpha\right)-cos\alpha}\)
\(a\in\left(\frac{\pi}{2};\pi\right)\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\)
\(A=\frac{sin\left(4\pi-\frac{\pi}{2}-a\right)}{sin\left(a+\frac{\pi}{4}\right)-cosa}=\frac{-sin\left(a+\frac{\pi}{2}\right)}{sin\left(a+\frac{\pi}{4}\right)-cosa}=\frac{-cosa}{sina.cos\frac{\pi}{4}+cosa.sin\frac{\pi}{4}-cosa}\)
\(=\frac{-\frac{4}{5}}{\frac{3}{5}.\frac{\sqrt{2}}{2}-\frac{4}{5}.\frac{\sqrt{2}}{2}-\frac{4}{5}}=...\)
Biết \(sin\alpha=\frac{3}{5}\) và \(\frac{\pi}{2}< \alpha< \pi\). Tính \(P=1-2sin^2\left(\frac{\pi}{4}-\alpha\right)+sin2\alpha+cos\left(\pi-2\alpha\right)-6tan\left(\frac{\pi}{2}-\alpha\right)\)
\(\frac{\pi}{2}< a< \pi\Rightarrow cosa< 0\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\frac{4}{5}\)
\(P=1-\left[1-cos\left(\frac{\pi}{2}-2a\right)\right]+sin2a-cos2a-6cota\)
\(=sin2a+sin2a-cos2a-6cota\)
\(=2sin2a-cos2a-6cota\)
\(=4sina.cosa-\left(cos^2a-sin^2a\right)-\frac{6cosa}{sina}\) (thay số và bấm máy)
Cho \(\sin\alpha+\cos\alpha=\frac{\sqrt{6}}{2},a\in\left(0;\frac{\pi}{4}\right)\)
Tính giá trị biểu thức: \(P=\cos\left(\alpha+\frac{\pi}{4}\right)+\sqrt{2\left(1-\sin\alpha\cos\alpha+\sin\alpha-\cos\alpha\right)}\)
