Cho a thỏa mãn tan a \( = \frac{{ - 4}}{3}\), a thuộc \(\left( {\frac{{3{\rm{\pi }}}}{2};\;\,2{\rm{\pi }}} \right)\).

Tính \({\rm{P}} = \sin \left( {\frac{{\rm{a}}}{2}} \right) + \cos \left( {\frac{{\rm{a}}}{2}} \right)\).

Lời giải:

Vì a \( \in \left( {\frac{{3{\rm{\pi }}}}{2};\;\,2{\rm{\pi }}} \right)\) nên \(\frac{{\rm{a}}}{2} \in \left( {\frac{{3{\rm{\pi }}}}{4};\;\,{\rm{\pi }}} \right)\).

Do đó, sin a < 0, cos a > 0, \(\sin \left( {\frac{{\rm{a}}}{2}} \right) > 0,\;\,\cos \left( {\frac{{\rm{a}}}{2}} \right) < 0\).

+) 1 + tan² a = 1 + \(\frac{{{{\sin }^2}{\rm{a}}}}{{{{\cos }^2}{\rm{a}}}} = \frac{{{{\sin }^2}{\rm{a}} + {{\cos }^2}{\rm{a}}}}{{{{\cos }^2}{\rm{a}}}} = \frac{1}{{{{\cos }^2}{\rm{a}}}}\)\( \Rightarrow \cos {\rm{a}} = \sqrt {\frac{1}{{1 + {{\tan }^2}{\rm{a}}}}} = \frac{3}{5}\).

Do đó, cos a = 2cos² \(\left( {\frac{{\rm{a}}}{2}} \right)\) – 1. Suy ra cos\(\left( {\frac{{\rm{a}}}{2}} \right)\)\( = - \sqrt {\frac{{1 + \cos {\rm{a}}}}{2}} = - \frac{{2\sqrt 5 }}{5}\).

+) \({\sin ^2}\left( {\frac{{\rm{a}}}{2}} \right) + {\cos ^2}\left( {\frac{{\rm{a}}}{2}} \right) = 1\) nên sin \(\left( {\frac{{\rm{a}}}{2}} \right)\)\( = \sqrt {1 - {{\cos }^2}\left( {\frac{{\rm{a}}}{2}} \right)} = \frac{{\sqrt 5 }}{5}\).

Vậy \({\rm{P}} = \sin \left( {\frac{{\rm{a}}}{2}} \right) + \cos \left( {\frac{{\rm{a}}}{2}} \right) = - \frac{{\sqrt 5 }}{5}\).