Bài tập 49 trang 215 SGK Toán 10 NC

a)

\(\begin{array}{*{20}{l}}
\begin{array}{l}
{\cos ^2}\left( {\alpha  + x} \right) + {\cos ^2}x\\
 - 2\cos \alpha \cos x.\cos \left( {\alpha  + x} \right)
\end{array}\\
\begin{array}{l}
 = \cos \left( {\alpha  + x} \right)[\cos \left( {\alpha  + x} \right)\\
 - 2\cos \alpha \cos x] + {\cos ^2}x
\end{array}\\
\begin{array}{l}
 = \cos \left( {\alpha  + x} \right)( - \cos \alpha \cos x\\
 - \sin \alpha \sin x) + {\cos ^2}x
\end{array}\\
{ =  - \cos \left( {\alpha  + x} \right)\cos \left( {\alpha  - x} \right) + {{\cos }^2}x}\\
{ =  - \frac{1}{2}\left( {\cos 2\alpha  + \cos 2x} \right) + {{\cos }^2}x}\\
\begin{array}{l}
 =  - \frac{1}{2}\cos 2\alpha  - \frac{{\cos 2x}}{2} + {\cos ^2}x\\
 =  - \frac{1}{2}\cos 2\alpha  + \frac{1}{2} = {\sin ^2}\alpha 
\end{array}
\end{array}\)

b)

\(\begin{array}{*{20}{l}}
{\sin 4x.\sin 10x - \sin 11x.\sin 3x - \sin 7x.{\rm{sinx}}}\\
\begin{array}{l}
{\rm{ = }}\frac{1}{2}\left( {\cos 6x - \cos 14x} \right) - \frac{1}{2}\left( {\cos 8x - \cos 14x} \right)\\
 - \frac{1}{2}\left( {\cos 6x - \cos 8x} \right) = 0
\end{array}
\end{array}\)

-- Mod Toán 10 HỌC247