Giải bài 25 trang 15 sách bài tập toán 11 - Cánh diều

Đề bài

Cho \(\sin a = \frac{2}{3}\) với \(\frac{\pi }{2} < a < \pi \). Tính:

a)    \(\cos a\), \(\tan a\)

b)    \(\sin \left( {a + \frac{\pi }{4}} \right)\), \(\cos \left( {a - \frac{{5\pi }}{6}} \right)\), \(\tan \left( {a + \frac{{2\pi }}{3}} \right)\)

c)     \(\sin 2a\), \(\cos 2a\)

Lời giải chi tiết

a) Ta có \({\sin ^2}a + {\cos ^2}a = 1 \Rightarrow {\cos ^2}a = 1 - {\sin ^2}a = 1 - {\left( {\frac{2}{3}} \right)^2} = \frac{5}{9} \Rightarrow \cos a =  \pm \frac{{\sqrt 5 }}{3}\).

Do \(\frac{\pi }{2} < a < \pi  \Rightarrow \cos a < 0 \Rightarrow \cos a =  - \frac{{\sqrt 5 }}{3}\).

Suy ra \(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{2}{3} :\frac{{ - \sqrt 5 }}{3} = \frac{{ - 2\sqrt 5 }}{5}\)

b) Ta có:

\(\sin \left( {a + \frac{\pi }{4}} \right) = \sin a\cos \frac{\pi }{4} + \cos a\sin \frac{\pi }{4} = \frac{2}{3}.\frac{{\sqrt 2 }}{2} + \frac{{ - \sqrt 5 }}{3}.\frac{{\sqrt 2 }}{2} = \frac{{2\sqrt 2  - \sqrt {10} }}{6}\)

\(\cos \left( {a - \frac{{5\pi }}{6}} \right) = \cos a\cos \frac{{5\pi }}{6} + \sin a\sin \frac{{5\pi }}{6} = \frac{{ - \sqrt 5 }}{3}.\frac{{ - \sqrt 3 }}{2} + \frac{2}{3}.\frac{1}{2} = \frac{{2 + \sqrt {15} }}{6}\)

\(\tan \left( {a + \frac{{2\pi }}{3}} \right) = \frac{{\tan a + \tan \frac{{2\pi }}{3}}}{{1 - \tan a\tan \frac{{2\pi }}{3}}} = \frac{{\frac{{ - 2\sqrt 5 }}{5} + \left( { - \sqrt 3 } \right)}}{{1 - \frac{{ - 2\sqrt 5 }}{5}.\left( { - \sqrt 3 } \right)}} = \frac{{8\sqrt 5  + 9\sqrt 3 }}{7}\)

c) \(\begin{array}{l}\sin 2a = 2\sin a.\cos a = 2.\frac{2}{3}.\left( { - \frac{{\sqrt 5 }}{3}} \right) =  - \frac{{4\sqrt 5 }}{9}\\\cos 2a = 1 - 2{\sin ^2}a = 1 - 2.{\left( {\frac{2}{3}} \right)^2} = \frac{1}{9}\end{array}\)