Bài tập 40 trang 213 SGK Toán 10 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sqrt 2 \sin \left( {\alpha  + \frac{\pi }{4}} \right)\\
 = \sqrt 2 \left( {\sin \alpha .\cos \frac{\pi }{4} + \sin \frac{\pi }{4}.\cos \alpha } \right)
\end{array}\\
\begin{array}{l}
 = \sqrt 2 \left( {\sin \alpha .\frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}\cos \alpha } \right)\\
 = \sin \alpha  + \cos \alpha 
\end{array}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sqrt 2 \sin \left( {\alpha  - \frac{\pi }{4}} \right)\\
 = \sqrt 2 \left( {\sin \alpha .\cos \frac{\pi }{4} - \sin \frac{\pi }{4}.\cos \alpha } \right)
\end{array}\\
\begin{array}{l}
 = \sqrt 2 \left( {\sin \alpha .\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}\cos \alpha } \right)\\
 = \sin \alpha  - \cos \alpha 
\end{array}
\end{array}\)

c)

\(\begin{array}{l}
\tan \left( {\frac{\pi }{4} - \alpha } \right) = \frac{{\tan \frac{\pi }{4} - \tan \alpha }}{{1 + \tan \frac{\pi }{4}.\tan \alpha }}\\
 = \frac{{1 - \tan \alpha }}{{1 + \tan \alpha }}
\end{array}\)

d)

\(\begin{array}{l}
\tan \left( {\frac{\pi }{4} + \alpha } \right) = \frac{{\tan \frac{\pi }{4} + \tan \alpha }}{{1 - \tan \frac{\pi }{4}.\tan \alpha }}\\
 = \frac{{1 + \tan \alpha }}{{1 - \tan \alpha }}
\end{array}\)

-- Mod Toán 10 HỌC247