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Bài tập 33 trang 206 SGK Toán 10 NC

Tính

a) Tính \(\sin \frac{{25\pi }}{6} + \cos \frac{{25\pi }}{3} + \tan \left( { - \frac{{25\pi }}{4}} \right)\)

b) Biết \(\sin \left( {\pi  + \alpha } \right) =  - \frac{1}{3}\), hãy tính \(\cos \left( {2\pi  - \alpha } \right)\), \(\tan \left( {\alpha  - 7\pi } \right)\) và \(\sin \left( {\frac{{3\pi }}{2} - \alpha } \right)\)

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin \frac{{25\pi }}{6} = \sin \left( {4\pi  + \frac{\pi }{6}} \right)\\
 = \sin \frac{\pi }{6} = \frac{1}{2}
\end{array}\\
\begin{array}{l}
\cos \frac{{25\pi }}{3} = \cos \left( {8\pi  + \frac{\pi }{3}} \right)\\
 = \cos \frac{\pi }{3} = \frac{1}{2}
\end{array}\\
\begin{array}{l}
\tan \left( { - \frac{{25\pi }}{4}} \right) =  - \tan \left( {6\pi  + \frac{\pi }{4}} \right)\\
 =  - \tan \frac{\pi }{4} =  - 1
\end{array}\\
{ \Rightarrow \sin \frac{{25\pi }}{6} + \cos \frac{{25\pi }}{3} + \tan \left( { - \frac{{25\pi }}{4}} \right) = 0}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
{\sin \left( {\pi  + \alpha } \right) =  - \frac{1}{3} \Rightarrow \sin \alpha  = \frac{1}{3}}\\
\begin{array}{l}
\cos \left( {2\pi  - \alpha } \right) = \cos \left( { - \alpha } \right) = \cos \alpha \\
 =  \pm \sqrt {1 - {{\sin }^2}\alpha }  =  \pm \frac{{2\sqrt 2 }}{3}
\end{array}\\
{\tan \left( {\alpha  - 7\pi } \right) = \tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} =  \pm \frac{1}{{2\sqrt 2 }}}\\
\begin{array}{l}
\sin \left( {\frac{{3\pi }}{2} - \alpha } \right) = \sin \left( {\pi  + \frac{\pi }{2} - \alpha } \right)\\
 =  - \sin \left( {\frac{\pi }{2} - \alpha } \right) =  - \cos \alpha  =  \pm \frac{{2\sqrt 2 }}{3}
\end{array}
\end{array}\)

 

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