a) Cho góc α thỏa mãn 3pi/ 2 < α < 2pi ; sin α = − 12 13 . Tính giá trị lượng giác cos ( pi/ 3 + α ) .

a) Vì \[\frac{{3\pi }}{2} < \alpha  < 2\pi \alpha \]\[\] nên \[\cos \alpha  > 0\].

Ta có: \[{\sin ^2}\alpha  + co{s^2}\alpha  = 1\].

Suy ra: \[\cos \alpha  = \sqrt {1 - {{\sin }^2}\alpha }  = \frac{5}{{13}}\].

Vậy \[cos\left( {\frac{\pi }{3} + \alpha } \right) = \cos \frac{\pi }{3}\cos \alpha  - \sin \frac{\pi }{3}\sin \alpha  = \frac{{5 + 12\sqrt 3 }}{{26}}\].

b) Ta có

\[\begin{array}{l}x\left( t \right) = {x_1}\left( t \right) + {x_2}\left( t \right) = 2\sqrt 3 \sin \left( {4\pi t + \frac{\pi }{6}} \right) + 2\cos \left( {4\pi t + \frac{\pi }{6}} \right)\\{\rm{                               }} = 4\left[ {\frac{{\sqrt 3 }}{2}\sin \left( {4\pi t + \frac{\pi }{6}} \right) + \frac{1}{2}\cos \left( {4\pi t + \frac{\pi }{6}} \right)} \right]\\{\rm{                               }} = 4\sin \left( {4\pi t + \frac{\pi }{3}} \right)\end{array}\]

 \[x\left( t \right) = 4\sin \left( {4\pi t + \frac{\pi }{3}} \right)\] có dạng \[x\left( t \right) = A\cos \left( {\omega t + \varphi } \right)\] do đó \(A = 4,\omega  = 4\pi ,\varphi  = \frac{\pi }{3}\)