a) Cho sin a = − 1/3 với a ∈ ( pi ; 3pi/ 2 ) . Tính giá trị của cos a , sin ( a + pi/ 3 ) .

a) Ta có \({\sin ^2}a + {\cos ^2}a = 1 \Rightarrow \cos a =  \pm \frac{{2\sqrt 2 }}{3}\).

Do \(a \in \left( {\pi \,;\,\frac{{3\pi }}{2}} \right)\) nên nhận \(\cos a =  - \frac{{2\sqrt 2 }}{3}\).

\(\sin \left( {a + \frac{\pi }{3}} \right) = \sin a\cos \frac{\pi }{3} + \cos a\sin \frac{\pi }{3}\).

\( = \left( { - \frac{1}{3}} \right).\frac{1}{2} + \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 3 }}{2} =  - \frac{{1 + 2\sqrt 6 }}{6}\).

b)Ta có \[A = \frac{{\left( {\sin 7x + \sin x} \right) + \sin 4x}}{{\left( {\cos 7x + \cos x} \right) + \cos 4x}} = \frac{{2\sin 4x.\cos 3x + \sin 4x}}{{2\cos 4x.\cos 3x + \cos 4x}}\]

                                                                   \[ = \frac{{\sin 4x.\left( {2\cos 3x + 1} \right)}}{{\cos 4x.\left( {2\cos 3x + 1} \right)}} = \tan 4x\]

c) Ta có \[2\sin \left( {2x - \frac{\pi }{3}} \right) =  - \sqrt 3 \] \[ \Leftrightarrow \sin \left( {2x - \frac{\pi }{3}} \right) = \sin \left( { - \frac{\pi }{3}} \right)\]

\[ \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} =  - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi  - \left( { - \frac{\pi }{3}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\]