Bài 2 trang 15 SGK Toán 11 tập 1 – Cánh Diều

Đề bài

Tính các giá trị lượng giác của mỗi góc sau: \(225^\circ ; – 225^\circ ; – 1035^\circ \);\(\frac{{5\pi }}{3};\frac{{19\pi }}{2}; – \frac{{159\pi }}{4}\)

Lời giải chi tiết

\(\begin{array}{l}\cos \left( {{{225}^ \circ }} \right) = \cos \left( {{{180}^ \circ } + {{45}^ \circ }} \right) =  – \cos \left( {{{45}^ \circ }} \right) =  – \frac{{\sqrt 2 }}{2}\\\sin \left( {{{225}^ \circ }} \right) = \sin \left( {{{180}^ \circ } + {{45}^ \circ }} \right) =  – \sin \left( {{{45}^ \circ }} \right) =  – \frac{{\sqrt 2 }}{2}\\\tan \left( {225^\circ } \right) = \frac{{\sin \left( {{{225}^ \circ }} \right)}}{{\cos \left( {{{225}^ \circ }} \right)}} = 1\\\cot \left( {225^\circ } \right) = \frac{1}{{\tan \left( {225^\circ } \right)}} = 1\end{array}\)

\(\begin{array}{l}\cos \left( { – {{225}^ \circ }} \right) = \cos \left( {{{225}^ \circ }} \right) = \cos \left( {{{180}^ \circ } + {{45}^ \circ }} \right) =  – \cos \left( {{{45}^ \circ }} \right) =  – \frac{{\sqrt 2 }}{2}\\\sin \left( { – {{225}^ \circ }} \right) =  – \sin \left( {{{225}^ \circ }} \right) =  – \sin \left( {{{180}^ \circ } + {{45}^ \circ }} \right) = \sin \left( {{{45}^ \circ }} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( { – 225^\circ } \right) = \frac{{\sin \left( {{{225}^ \circ }} \right)}}{{\cos \left( {{{225}^ \circ }} \right)}} =  – 1\\\cot \left( { – 225^\circ } \right) = \frac{1}{{\tan \left( {225^\circ } \right)}} =  – 1\end{array}\)

\(\begin{array}{l}\cos \left( { – {{1035}^ \circ }} \right) = \cos \left( {{{1035}^ \circ }} \right) = \cos \left( {{{6.360}^ \circ } – {{45}^ \circ }} \right) = \cos \left( { – {{45}^ \circ }} \right) = \cos \left( {{{45}^ \circ }} \right) = \frac{{\sqrt 2 }}{2}\\\sin \left( { – {{1035}^ \circ }} \right) =  – \sin \left( {{{1035}^ \circ }} \right) =  – \sin \left( {{{6.360}^ \circ } – {{45}^ \circ }} \right) =  – \sin \left( { – {{45}^ \circ }} \right) = \sin \left( {{{45}^ \circ }} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( { – 1035^\circ } \right) = \frac{{\sin \left( { – {{1035}^ \circ }} \right)}}{{\cos \left( { – {{1035}^ \circ }} \right)}} = 1\\\cot \left( { – 1035^\circ } \right) = \frac{1}{{\tan \left( { – 1035^\circ } \right)}} =  – 1\end{array}\)

\(\begin{array}{l}\cos \left( {\frac{{5\pi }}{3}} \right) = \cos \left( {\pi  + \frac{{2\pi }}{3}} \right) =  – \cos \left( {\frac{{2\pi }}{3}} \right) = \frac{1}{2}\\\sin \left( {\frac{{5\pi }}{3}} \right) = \sin \left( {\pi  + \frac{{2\pi }}{3}} \right) =  – \sin \left( {\frac{{2\pi }}{3}} \right) =  – \frac{{\sqrt 3 }}{2}\\\tan \left( {\frac{{5\pi }}{3}} \right) = \frac{{\sin \left( {\frac{{5\pi }}{3}} \right)}}{{\cos \left( {\frac{{5\pi }}{3}} \right)}} =  – \sqrt 3 \\\cot \left( {\frac{{5\pi }}{3}} \right) = \frac{1}{{\tan \left( {\frac{{5\pi }}{3}} \right)}} =  – \frac{{\sqrt 3 }}{3}\end{array}\)

\(\begin{array}{l}\cos \left( {\frac{{19\pi }}{2}} \right) = \cos \left( {8\pi  + \frac{{3\pi }}{2}} \right) = \cos \left( {\frac{{3\pi }}{2}} \right) = \cos \left( {\pi  + \frac{\pi }{2}} \right) =  – \cos \left( {\frac{\pi }{2}} \right) = 0\\\sin \left( {\frac{{19\pi }}{2}} \right) = \sin \left( {8\pi  + \frac{{3\pi }}{2}} \right) = \sin \left( {\frac{{3\pi }}{2}} \right) = \sin \left( {\pi  + \frac{\pi }{2}} \right) =  – \sin \left( {\frac{\pi }{2}} \right) =  – 1\\\tan \left( {\frac{{19\pi }}{2}} \right)\\\cot \left( {\frac{{19\pi }}{2}} \right) = \frac{{\cos \left( {\frac{{19\pi }}{2}} \right)}}{{\sin \left( {\frac{{19\pi }}{2}} \right)}} = 0\end{array}\)

\(\begin{array}{l}\cos \left( { – \frac{{159\pi }}{4}} \right) = \cos \left( {\frac{{159\pi }}{4}} \right) = \cos \left( {40.\pi  – \frac{\pi }{4}} \right) = \cos \left( { – \frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\sin \left( { – \frac{{159\pi }}{4}} \right) =  – \sin \left( {\frac{{159\pi }}{4}} \right) =  – \sin \left( {40.\pi  – \frac{\pi }{4}} \right) =  – \sin \left( { – \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\\tan \left( { – \frac{{159\pi }}{4}} \right) = \frac{{\cos \left( { – \frac{{159\pi }}{4}} \right)}}{{\sin \left( { – \frac{{159\pi }}{4}} \right)}} = 1\\\cot \left( { – \frac{{159\pi }}{4}} \right) = \frac{1}{{\tan \left( { – \frac{{159\pi }}{4}} \right)}} = 1\end{array}\)