Bạn ơi sin gì đó bạn ?

a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\)

Ta có: \({\sin ^2}a + {\cos ^2}a  = 1\)

 \(\Leftrightarrow \frac{1}{9} + {\cos ^2}a  = 1\)

\(\Leftrightarrow {\cos ^2}a =  1 - \frac{1}{9}= \frac{8}{9}\)

\(\Leftrightarrow \cos a  =\pm\sqrt { \frac{8}{9}}  =  \pm \frac{{2\sqrt 2 }}{3}\)

Vì \(\cos a < 0\) nên \(cos a =-\frac{{2\sqrt 2 }}{3}\)

Suy ra \(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} =  - \frac{{\sqrt 2 }}{4}\)

Ta có: \(\sin 2a = 2\sin a\cos a = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) =  - \frac{{4\sqrt 2 }}{9}\)

\(\cos 2a = 1 - 2{\sin ^2}a = 1 - \frac{2}{9} = \frac{7}{9}\)

\(\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}} = \frac{{2.\left( { - \frac{{\sqrt 2 }}{4}} \right)}}{{1 - {{\left( { - \frac{{\sqrt 2 }}{4}} \right)}^2}}} =  - \frac{{4\sqrt 2 }}{7}\)

b) Vì \(\frac{\pi }{2} < a < \frac{{3\pi }}{4}\) nên \(\sin a > 0,\cos a < 0\)

\({\left( {\sin a + \cos a} \right)^2} = {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = 1 + 2\sin a\cos a = \frac{1}{4}\)

Suy ra \(\sin 2a = 2\sin a\cos a = \frac{1}{4} - 1 =  - \frac{3}{4}\)

Ta có: \({\sin ^2}a + {\cos ^2}a = 1\;\)

\( \Leftrightarrow \left( {\frac{1}{2} - {\cos }a} \right)^2 + {\cos ^2}a - 1 = 0\)

\( \Leftrightarrow \frac{1}{4} - \cos a + {\cos ^2}a + {\cos ^2}a - 1 = 0\)

\( \Leftrightarrow 2{\cos ^2}a - \cos a - \frac{3}{4} = 0\)

\( \Rightarrow \cos a = \frac{{1 - \sqrt 7 }}{4}\) (Vì \(\cos a < 0)\)

\(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{{1 - \sqrt 7 }}{4}} \right)^2} - 1 =  - \frac{{\sqrt 7 }}{4}\)

\(\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{3}{4}}}{{ - \frac{{\sqrt 7 }}{4}}} = \frac{{3\sqrt 7 }}{7}\)

Ta có:

 \(\begin{array}{l}\sin \left( { - \frac{{15\pi }}{2} - \alpha } \right) - \cos \left( {13\pi  + \alpha } \right) =  \sin \left( { -\frac{{16\pi }}{2} +\frac{{\pi }}{2}  + \alpha } \right) - \cos \left( {12\pi  + \pi + \alpha } \right) =  \sin \left( {-8\pi  + \frac{\pi }{2} - \alpha } \right) - \cos \left( { \pi + \alpha } \right) \\ = \sin \left( {\frac{\pi }{2} - \alpha } \right) + \cos \left( \alpha  \right) = \cos \left( \alpha  \right) + \cos \left( \alpha  \right) = 2\cos \left( \alpha  \right) = 2.\left( { - \frac{5}{{13}}} \right) = \frac{{ - 10}}{{13}}\end{array}\)

\(A = \cos {75^0}\cos {15^0} = \frac{1}{2}\left[ {\cos \left( {{{75}^0} - {{15}^0}} \right) + \cos \left( {{{75}^0} + {{15}^0}} \right)} \right] \\= \frac{1}{2}.\cos {60^0}.\cos {90^0} = 0\)

\(B = \sin \frac{{5\pi }}{{12}}\cos \frac{{7\pi }}{{12}} = \frac{1}{2}\left[ {\sin \left( {\frac{{5\pi }}{{12}} - \frac{{7\pi }}{{12}}} \right) + \sin \left( {\frac{{5\pi }}{{12}} + \frac{{7\pi }}{{12}}} \right)} \right] \\= \frac{1}{2}\sin \left( { - \frac{{2\pi }}{{12}}} \right).\sin \left( {\frac{{12\pi }}{{12}}} \right) =  - \frac{1}{2}\sin \frac{\pi }{6}\sin \pi  = 0\)

Áp dụng công thức cộng, ta có:

\(\begin{array}{l}\sin \frac{\pi }{{12}} = \sin \left( {\frac{\pi }{4} - \frac{\pi }{6}} \right) = \sin \frac{\pi }{4}.\cos \frac{\pi }{6} - \cos \frac{\pi }{4}.\sin \frac{\pi }{6}\\ = \frac{{\sqrt 2 }}{2}.\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 2 }}{2}.\frac{1}{2} = \frac{{\sqrt 6  - \sqrt 2 }}{4}\end{array}\)

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

\(\begin{array}{l}Q = {\tan ^2}\frac{\pi }{3} + {\sin ^2}\frac{\pi }{4} + \cot \frac{\pi }{4} + \cos \frac{\pi }{2}\\\,\,\,\,\, = \,{\left( {\sqrt 3 } \right)^2} + {\left( {\frac{{\sqrt 2 }}{2}} \right)^2} + 1 + 0 = \frac{7}{2}\end{array}\)

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) Ta có:

      \(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x.\frac{{\sqrt 2 }}{2} + \cos x.\frac{{\sqrt 2 }}{2}} \right) = \sin x + \cos x\)

b) Ta có:

\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)