Bài tập 1 trang 153 SGK Đại số 10

Bài tập 1 trang 153 SGK Đại số 10

Câu a:

* \(\cos {225^0} = \cos ({270^0} - {45^0}) = \cos ({180^0} + {90^0} - {45^0})\)

\( =  - \cos ({90^0} - {45^0}) =  - \sin {45^0} =  - \frac{{\sqrt 2 }}{2}.\)

* \(\sin {240^0} = \sin ({180^0} + {90^0} - {30^0}) =  - \sin ({90^0} - {30^0})\)

\( =  - \cos {30^0} =  - \frac{{\sqrt 3 }}{2}\)

* \(\cot ( - {15^0}) =  - \cot {15^0} =  - \cot ({45^0} - {30^0}) = \frac{{ - 1}}{{\tan ({{45}^0} - {{30}^0})}}\)

\( =  - \frac{{1 + \tan {{45}^0}\tan {{30}^0}}}{{\tan {{45}^0} - \tan {{30}^0}}} = \frac{{1 + \frac{1}{{\sqrt 3 }}}}{{1 - \frac{1}{{\sqrt 3 }}}} =  - \frac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}}\)

\( =  - \frac{{1 - \sqrt 3 }}{{1 - \sqrt 3 }} = \frac{{{{(1 + \sqrt 3 )}^2}}}{2} =  - \frac{{4 + 2\sqrt 3 }}{2} =  - 2 - \sqrt 3 \)

* \(\tan {75^0} = \tan ({45^0} + {30^0}) = \frac{{\tan {{45}^0} + \tan {{30}^0}}}{{1 - \tan {{45}^0}.\tan {{30}^0}}}\)

\( = \frac{{1 + \frac{1}{{\sqrt 3 }}}}{{1 - \frac{1}{{\sqrt 3 }}}} = \frac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}} = \frac{{{{\left( {\sqrt 3  + 1} \right)}^2}}}{2}\)

Câu b:

* \(\sin \frac{{7\pi }}{{12}} = \sin \left( {\frac{{3\pi }}{{12}} + \frac{{4\pi }}{{12}}} \right) = \sin \left( {\frac{\pi }{3} + \frac{\pi }{4}} \right) = \sin \frac{\pi }{3}.\cos \frac{\pi }{4} + \cos \frac{\pi }{3}.\sin \frac{\pi }{4}\)

\( = \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 2 }}{2} + \frac{1}{2}.\frac{{\sqrt 2 }}{2} = \frac{{\sqrt 6  - \sqrt 2 }}{4}\)

* \(\cos \left( {\frac{{ - \pi }}{{12}}} \right) = \cos \left( {\frac{\pi }{4} - \frac{\pi }{3}} \right) = \cos \frac{\pi }{4}.\cos \frac{\pi }{3} + \sin \frac{\pi }{4}.\sin \frac{\pi }{3}\)

\( = \frac{{\sqrt 2 }}{2}.\frac{1}{2} + \frac{{\sqrt 2 }}{2}.\frac{{\sqrt 3 }}{2} = \frac{{\sqrt 2  + \sqrt 6 }}{4}\)

* \(\tan \frac{{13\pi }}{{12}} = \tan \left( {\pi  + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{{12}} = \tan \left( {\frac{\pi }{3} - \frac{\pi }{4}} \right)\)

\( = \frac{{\tan \frac{\pi }{3} - \tan \frac{\pi }{4}}}{{1 + \tan \frac{\pi }{3}.\tan \frac{\pi }{4}}} = \frac{{\sqrt 3  - 1}}{{1 + \sqrt 3 }} = \frac{{{{\left( {\sqrt 3  - 1} \right)}^2}}}{2} = 2 - \sqrt 3 \)

-- Mod Toán 10