Giải Bài 1.9 trang 21 SGK Toán 11 Kết nối tri thức tập 1 - KNTT

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

Do đó: \(\cos a = \sqrt {1 - {{\sin }^2}a}  = \sqrt {1 - \frac{1}{9}}  =  - \frac{{2\sqrt 2 }}{3}\)

Suy ra

\(\begin{array}{l} \tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} = - \frac{{\sqrt 2 }}{4}\\ \sin 2a = 2\sin a\cos a = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) = - \frac{{4\sqrt 2 }}{9}\\ \cos 2a = 1 - 2{\sin ^2}a = 1 - \frac{1}{9} = \frac{8}{9}\\ \tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}} = \frac{{2.\left( { - \frac{{\sqrt 2 }}{4}} \right)}}{{1 - {{\left( { - \frac{{\sqrt 2 }}{4}} \right)}^2}}} = - \frac{{4\sqrt 2 }}{7} \end{array}\)

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

\(\begin{array}{l} {\sin ^2}a + {\cos ^2}a = 1\:\\ \Leftrightarrow \left( {\frac{1}{2} - {{\cos }^2}a} \right) + {\cos ^2}a - 1 = 0\\ \Leftrightarrow \frac{1}{4} - \cos a + {\cos ^2}a + {\cos ^2}a - 1 = 0\\ \Leftrightarrow 2{\cos ^2}a - \cos a - \frac{3}{4} = 0\\ \Rightarrow \cos a = \frac{{1 - \sqrt 7 }}{4}\\ \cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{{1 - \sqrt 7 }}{4}} \right)^2} - 1 = - \frac{{\sqrt 7 }}{4}\\ \tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{3}{4}}}{{ - \frac{{\sqrt 7 }}{4}}} = \frac{{3\sqrt 7 }}{7} \end{array}\)