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Bài tập 20 trang 201 SGK Toán 10 NC

Bài tập 20 trang 201 SGK Toán 10 NC

Tính các giá trị lượng giác của các góc sau:

\(\begin{array}{l}
{225^0}; - {225^0};{750^0}; - {510^0}\\
;\frac{{5\pi }}{3};\frac{{11\pi }}{6};\frac{{ - 10\pi }}{3}; - \frac{{17\pi }}{3}
\end{array}\)

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Hướng dẫn giải chi tiết

\(\begin{array}{*{20}{l}}
\begin{array}{l}
*\sin {225^0} = \sin \left( {{{45}^0} + {{180}^0}} \right)\\
 =  - \sin {45^0} =  - \frac{{\sqrt 2 }}{2}
\end{array}\\
\begin{array}{l}
\cos {225^0} = \cos \left( {{{45}^0} + {{180}^0}} \right)\\
 =  - \cos {45^0} =  - \frac{{\sqrt 2 }}{2}
\end{array}\\
{\tan {{225}^0} = \cot {{225}^0} = 1}
\end{array}\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
*\sin \left( { - {{225}^0}} \right) = \sin \left( { - {{45}^0} - {{180}^0}} \right)\\
 = \sin {45^0} = \frac{{\sqrt 2 }}{2}
\end{array}\\
\begin{array}{l}
\cos \left( { - {{225}^0}} \right) = \cos \left( { - {{45}^0} - {{180}^0}} \right)\\
 = \cos \left( {{{45}^0} + {{180}^0}} \right) =  - \frac{{\sqrt 2 }}{2}
\end{array}\\
{\tan \left( { - {{225}^0}} \right) = \cot \left( { - {{225}^0}} \right) =  - 1}
\end{array}\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
*\sin {750^0} = \sin \left( {{{30}^0} + {{2.360}^0}} \right)\\
 = \sin {30^0} = \frac{1}{2}
\end{array}\\
\begin{array}{l}
\cos {750^0} = \cos \left( {{{30}^0} + {{2.360}^0}} \right)\\
 = \cos {30^0} = \frac{{\sqrt 3 }}{2}
\end{array}\\
{\tan {{750}^0} = \frac{{\sin {{750}^0}}}{{\cos {{750}^0}}} = \frac{{\sqrt 3 }}{3}}\\
{\cot {{750}^0} = \sqrt 3 }
\end{array}\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
*\sin \left( { - {{510}^0}} \right) = \sin \left( { - {{150}^0} - {{360}^0}} \right)\\
 = \sin \left( { - {{150}^0}} \right) =  - \frac{1}{2}
\end{array}\\
{\cos \left( { - {{510}^0}} \right) = \cos \left( { - {{150}^0}} \right) =  - \frac{{\sqrt 3 }}{2}}\\
{\tan \left( { - {{510}^0}} \right) = \frac{{\sqrt 3 }}{3}}\\
{\cot \left( { - {{510}^0}} \right) = \sqrt 3 }
\end{array}\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
*\sin \frac{{5\pi }}{3} = \sin \left( { - \frac{\pi }{3} + 2\pi } \right)\\
 = \sin \left( { - \frac{\pi }{3}} \right) =  - \frac{{\sqrt 3 }}{2}
\end{array}\\
{\cos \frac{{5\pi }}{3} = \cos \left( { - \frac{\pi }{3}} \right) = \frac{1}{2}}\\
{\tan \frac{{5\pi }}{3} =  - \sqrt 3 }\\
{\cot \frac{{5\pi }}{3} =  - \frac{1}{{\sqrt 3 }}}
\end{array}\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
*\sin \frac{{11\pi }}{6} = \sin \left( { - \frac{\pi }{6} + 2\pi } \right)\\
 = \sin \left( { - \frac{\pi }{6}} \right) = \sin \frac{\pi }{6} =  - \frac{1}{2}
\end{array}\\
{\cos \frac{{11\pi }}{6} = \cos \left( { - \frac{\pi }{6}} \right) = \cos \frac{\pi }{6} = \frac{{\sqrt 3 }}{2}}\\
{\tan \frac{{11\pi }}{6} =  - \frac{1}{{\sqrt 3 }}}\\
{\cot \frac{{11\pi }}{6} =  - \sqrt 3 }
\end{array}\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
*\sin \left( { - \frac{{10\pi }}{3}} \right) = \sin \left( {\frac{{2\pi }}{3} - 4\pi } \right)\\
 = \sin \frac{{2\pi }}{3} = \frac{{\sqrt 3 }}{2}
\end{array}\\
{\cos \left( { - \frac{{10\pi }}{3}} \right) = \cos \frac{{2\pi }}{3} =  - \frac{1}{2}}\\
{\tan \left( { - \frac{{10\pi }}{3}} \right) =  - \sqrt 3 }\\
{\cot \left( { - \frac{{10\pi }}{3}} \right) =  - \frac{1}{{\sqrt 3 }}}
\end{array}\)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
*\sin \left( { - \frac{{17\pi }}{3}} \right) = \sin \left( {\frac{\pi }{3} - 6\pi } \right)\\
 = \sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}
\end{array}\\
{\cos \left( { - \frac{{17\pi }}{3}} \right) = \cos \frac{\pi }{3} = \frac{1}{2}}\\
{\tan \left( { - \frac{{17\pi }}{3}} \right) = \sqrt 3 }\\
{\cot \left( { - \frac{{17\pi }}{3}} \right) = \frac{1}{{\sqrt 3 }}}
\end{array}\)

-- Mod Toán 10 HỌC247

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