Bài tập 1 trang 223 SGK Toán 11 NC

Bài tập 1 trang 223 SGK Toán 11 NC

a. Ta có:

\(\begin{array}{l}
{\sin ^2}\frac{\pi }{8} = \frac{{1 - \cos \frac{\pi }{4}}}{2} = \frac{{1 - \frac{{\sqrt 2 }}{2}}}{2} = \frac{{2 - \sqrt 2 }}{4}\\
 \Rightarrow \sin \frac{\pi }{8} = \frac{1}{2}\sqrt {2 - \sqrt 2 } \\
{\cos ^2}\frac{\pi }{8} = \frac{{1 + \cos \frac{\pi }{4}}}{2} = \frac{{1 + \frac{{\sqrt 2 }}{2}}}{2} = \frac{{2 + \sqrt 2 }}{4}\\
\cos \frac{\pi }{8} = \frac{1}{2}\sqrt {2 + \sqrt 2 } 
\end{array}\)

b. Ta có:

\({1^2} + {\left( {\sqrt 2  - 1} \right)^2} = 4 - 2\sqrt 2 \)

Do đó:

\(\begin{array}{l}
\sin x + \left( {\sqrt 2  - 1} \right)\cos x\\
 = \left( {\sqrt {4 - 2\sqrt 2 } } \right)\left( {\frac{1}{{\sqrt {4 - 2\sqrt 2 } }}\sin x + \frac{{\sqrt 2  - 1}}{{\sqrt {4 - 2\sqrt 2 } }}\cos x} \right)\\
 = \sqrt {4 - 2\sqrt 2 } \left( {\sin x\cos \frac{\pi }{8} + \sin \frac{\pi }{8}\cos x} \right)\\
 = \sqrt {4 - 2\sqrt 2 } \sin \left( {x + \frac{\pi }{8}} \right)\\
 = \sqrt {4 - 2\sqrt 2 } \cos \left( {x - \frac{{3\pi }}{8}} \right)
\end{array}\)

Vì \(\frac{1}{{\sqrt {4 - 2\sqrt 2 } }} = \frac{{\sqrt {4 + 2\sqrt 2 } }}{{\sqrt 8 }} = \frac{1}{2}\sqrt {2 + \sqrt 2 }  = \cos \frac{\pi }{8}\)

Vậy \(C = \sqrt {4 - 2\sqrt 2 } \)

-- Mod Toán 11