Bài 10 trang 161 SGK Đại số 10

LG a

\(\displaystyle \cos {x \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5}\)

Lời giải chi tiết:

Nhân biểu thức với \(\sin {x \over 5}\),ta có:

\(\eqalign{
& A\sin {x \over 5}\cr& = \sin {x \over 5}\cos {x \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr 
& = {1 \over 2}\sin {{2x} \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr 
& = {1 \over 4}\sin {{4x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr&= {1 \over 8}\sin {{8x} \over 5}\cos {{8x} \over 5} \cr 
& = {1 \over {16}}\sin {{16x} \over 5} \cr} \)

Suy ra biểu thức rút gọn \(A = \dfrac{{\sin \dfrac{{16x}}{5}}}{{16\sin \dfrac{x}{5}}}\)

LG b

\(\displaystyle  \sin {x \over 7} + 2\sin {{3x} \over 7} + \sin {{5x} \over 7}\)

Lời giải chi tiết:

\(\begin{array}{l}
B = \sin \dfrac{x}{7} + 2\sin \dfrac{{3x}}{7} + \sin \dfrac{{5x}}{7}\\
= \left( {\sin \dfrac{{5x}}{7} + \sin \dfrac{x}{7}} \right) + 2\sin \dfrac{{3x}}{7}\\
= 2\sin \dfrac{{\dfrac{{5x}}{7} + \dfrac{x}{7}}}{2}\cos \dfrac{{\dfrac{{5x}}{7} - \dfrac{x}{7}}}{2} + 2\sin \dfrac{{3x}}{7}\\
= 2\sin \dfrac{{3x}}{7}\cos \dfrac{{2x}}{7} + 2\sin \dfrac{{3x}}{7}\\
= 2\sin \dfrac{{3x}}{7}\left( {\cos \dfrac{{2x}}{7} + 1} \right)\\
= 2\sin \dfrac{{3x}}{7}\left( {2{{\cos }^2}\dfrac{x}{7} - 1 + 1} \right)\\
= 2\sin \dfrac{{3x}}{7}.2{\cos ^2}\dfrac{x}{7}\\
= 4\sin \dfrac{{3x}}{7}{\cos ^2}\dfrac{x}{7}
\end{array}\)

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