Bài tập 2 trang 153 SGK Đại số 10

Câu a:

Ta có \(\cos \alpha  = \sqrt {1 - {{\sin }^2}\alpha }  = \sqrt {1 - \frac{1}{3}} \)

\( = \sqrt {\frac{2}{3}} \)(do \(0 < \alpha  < \frac{\pi }{2}\) nên \(\cos \alpha  > 0\) )

Vậy

\(\begin{array}{l}\cos \left( {\alpha  + \frac{\pi }{3}} \right) = \cos \alpha .\cos \frac{\pi }{3} - \sin \alpha .sin\frac{\pi }{3}\\ = \sqrt {\frac{2}{3}} .\frac{1}{2} - \frac{1}{{\sqrt 3 }}.\frac{{\sqrt 3 }}{2} = \frac{1}{{\sqrt 6 }} - \frac{1}{2}\end{array}\)  

Câu b:

Áp dụng \(1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }}\)

\( \Rightarrow {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }} - 1 = \frac{1}{{\left( { - \frac{1}{3}} \right)}} - 1 = 9 - 1 = 8\)

\( \Rightarrow \tan \alpha  =  - 2\sqrt 2 \) (do \(\frac{\alpha }{2} < \alpha  < \pi \) nên \(\tan \alpha  < 0\))

\(\tan \left( {\alpha  - \frac{\pi }{4}} \right) = \frac{{\tan \alpha  - \tan \frac{\pi }{4}}}{{1 + \tan \alpha .\tan \frac{\pi }{4}}}\)

\( = \frac{{ - 2\sqrt 2  - 1}}{{1 - 2\sqrt 2 }} = \frac{{1 + 2\sqrt 2 }}{{2\sqrt 2  - 1}} = \frac{{{{(1 + 2\sqrt 2 )}^2}}}{7} = \frac{{9 + 4\sqrt 2 }}{7}\)

Câu c:

Do \({0^0} < a < {90^0} \Rightarrow c{\rm{osa = }}\sqrt {1 - {{\sin }^2}a}  = \sqrt {1 - {{\left( {\frac{4}{5}} \right)}^2}}  = \sqrt {1 - \frac{{16}}{{25}}}  = \frac{3}{5}\)

Do \({90^0} < b < {180^0} \Rightarrow c{\rm{osb = }}\sqrt {1 - {{\sin }^2}b}  =  - \sqrt {1 - \frac{4}{9}}  = \sqrt {\frac{5}{9}}  =  - \frac{{\sqrt 5 }}{3}\)

Vậy ta có:

\({\rm{cos(a + b) = }}\cos a.\cos b - \sin a.\sin b = \frac{3}{5}.\left( { - \frac{{\sqrt 5 }}{3}} \right) - \frac{4}{5}.\frac{2}{3} =  - \frac{1}{{\sqrt 5 }} - \frac{8}{{15}}\)

\({\rm{sin(a - b) = }}\sin a.\cos b - c{\rm{os}}a.\sin b = \frac{4}{5}.\left( { - \frac{{\sqrt 5 }}{3}} \right) - \frac{3}{5}.\frac{2}{3} = \frac{4}{{3\sqrt 5 }} - \frac{2}{5}\)

-- Mod Toán 10 HỌC247