Bài tập 56 trang 218 SGK Toán 10 NC

a) Ta có:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
 - \frac{\pi }{2} < \alpha  < 0 \Rightarrow \sin \alpha  < 0\\
 \Rightarrow \sin \alpha  =  - \sqrt {1 - {{\cos }^2}\alpha }  = \frac{3}{5}
\end{array}\\
{\sin 2\alpha  = 2\sin \alpha \cos \alpha  =  - \frac{{24}}{{25}}}\\
{\cos 2\alpha  = 2{{\cos }^2}\alpha  - 1 = \frac{7}{{25}}}\\
{\cos \frac{\alpha }{2} = \sqrt {\frac{{1 + \cos \alpha }}{2}}  = \frac{{3\sqrt {10} }}{{10}}}\\
{\sin \frac{\alpha }{2} =  - \sqrt {\frac{{1 - \cos \alpha }}{2}}  =  - \frac{{\sqrt {10} }}{{10}}}
\end{array}\)

b) Vì \(\pi  < \alpha  < \frac{{3\pi }}{2} \)

\(\Rightarrow \tan \alpha  > 0\)

Do đó, ta có:

\(\begin{array}{*{20}{l}}
{\tan \alpha  = \sqrt {\frac{1}{{{{\cos }^2}\alpha }} - 1}  = \frac{{2\sqrt {10} }}{9}}\\
\begin{array}{l}
\tan \left( {\frac{\pi }{4} - \alpha } \right) = \frac{{1 - \tan \alpha }}{{1 + \tan \alpha }}\\
 = \frac{{121 - 36\sqrt {10} }}{{41}}
\end{array}
\end{array}\)

c)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{\sin ^4}\alpha  - {\cos ^4}\alpha \\
 = \left( {{{\sin }^2}\alpha  - {{\cos }^2}\alpha } \right)\left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)
\end{array}\\
{ = {{\sin }^2}\alpha  - {{\cos }^2}\alpha  =  - \cos 2\alpha  =  - \frac{3}{5}}
\end{array}\)

d) Ta có:

\(\begin{array}{l}
{\left( {\sin \alpha  - \sin \beta } \right)^2} = {\left( {\frac{1}{3}} \right)^2}\\
 \Rightarrow {\sin ^2}\alpha  + {\sin ^2}\beta  - 2\sin \alpha \sin \beta  = \frac{1}{9}\,\,\left( 1 \right)\\
{\left( {\cos \alpha  - \cos \beta } \right)^2} = {\left( {\frac{1}{2}} \right)^2}\\
 \Rightarrow {\cos ^2}\alpha  + {\cos ^2}\beta  - 2\cos \alpha \cos \beta  = \frac{1}{4}\,\,\left( 2 \right)
\end{array}\)

Cộng theo vế của (1) và (2), ta được:

\(\begin{array}{*{20}{l}}
\begin{array}{l}
1 + 1 - 2\left( {\cos \alpha \cos \beta  + \sin \alpha \sin \beta } \right)\\
 = \frac{1}{9} + \frac{1}{4} = \frac{{13}}{{36}}
\end{array}\\
{ \Rightarrow \cos \left( {\alpha  - \beta } \right) = \frac{{59}}{{72}}}
\end{array}\)

e)

\(\begin{array}{l}
\sin \frac{\pi }{{16}}\sin \frac{{3\pi }}{{16}}\sin \frac{{5\pi }}{{16}}\sin \frac{{7\pi }}{{16}}\\
 = \sin \frac{\pi }{{16}}\sin \frac{{3\pi }}{{16}}\sin \left( {\frac{\pi }{2} - \frac{{3\pi }}{{16}}} \right)\sin \left( {\frac{\pi }{2} - \frac{\pi }{{16}}} \right)\\
 = \sin \frac{\pi }{{16}}\sin \frac{{3\pi }}{{16}}\cos \frac{{3\pi }}{{16}}\cos \frac{\pi }{{16}}\\
 = \frac{1}{2}\sin \frac{\pi }{8}.\frac{1}{2}\sin \frac{{3\pi }}{8}\\
 = \frac{1}{4}\sin \frac{\pi }{8}\sin \left( {\frac{\pi }{2} - \frac{\pi }{8}} \right)\\
 = \frac{1}{4}\sin \frac{\pi }{8}\cos \frac{\pi }{8} = \frac{1}{8}\sin \frac{\pi }{4} = \frac{{\sqrt 2 }}{{16}}
\end{array}\)

-- Mod Toán 10 HỌC247