Đáp án:
C
Giải thích các bước giải:
$\begin{array}{l}
\cos \alpha + \cos \left( {\alpha + \dfrac{\pi }{5}} \right) + \cos \left( {\alpha + \dfrac{{2\pi }}{5}} \right) + ... + \cos \left( {\alpha + \dfrac{{9\pi }}{5}} \right)\\
= \cos \alpha + \left[ {\cos \left( {\alpha + \dfrac{\pi }{5}} \right) + \cos \left( {\alpha + \dfrac{{9\pi }}{5}} \right)} \right]\\
+ ... + \left[ {\cos \left( {\alpha + \dfrac{{4\pi }}{5}} \right) + \cos \left( {\alpha + \dfrac{{6\pi }}{5}} \right)} \right] + \cos \left( {\alpha + \dfrac{{5\pi }}{5}} \right)\\
= \cos \alpha + 2\cos \left( {\dfrac{{\alpha + \dfrac{\pi }{5} + \alpha + \dfrac{{9\pi }}{5}}}{2}} \right)\cos \left( {\dfrac{{\alpha + \dfrac{\pi }{5} - \alpha - \dfrac{{9\pi }}{5}}}{2}} \right)\\
+ ... + 2\cos \left( {\dfrac{{\alpha + \dfrac{{4\pi }}{5} + \alpha + \dfrac{{6\pi }}{5}}}{2}} \right)\cos \left( {\dfrac{{\alpha + \dfrac{{4\pi }}{5} - \alpha - \dfrac{{6\pi }}{5}}}{2}} \right) + \cos \left( {\alpha + \pi } \right)\\
= \cos \alpha + 2\cos \left( {\alpha + \pi } \right)\cos \left( { - \dfrac{{4\pi }}{5}} \right) + ... + 2\cos \left( {\alpha + \pi } \right)\cos \left( { - \dfrac{\pi }{5}} \right) + \cos \left( {\alpha + \pi } \right)\\
= \cos \alpha - 2\cos \alpha \cos \dfrac{{4\pi }}{5} - 2\cos \alpha \cos \dfrac{{3\pi }}{5} - 2\cos \alpha \cos \dfrac{{2\pi }}{5} - 2\cos \alpha \cos \dfrac{\pi }{5} - \cos \alpha \\
= - 2\cos \alpha \left( {\cos \dfrac{{4\pi }}{5} + \cos \dfrac{{3\pi }}{5} + \cos \dfrac{{2\pi }}{5} + \cos \dfrac{\pi }{5}} \right)\\
= - 2\cos \alpha \left[ {\left( {\cos \dfrac{{4\pi }}{5} + \cos \dfrac{\pi }{5}} \right) + \left( {\cos \dfrac{{3\pi }}{5} + \cos \dfrac{{2\pi }}{5}} \right)} \right]\\
= - 2\cos \alpha \left[ {2\cos \dfrac{\pi }{2}\cos \dfrac{{3\pi }}{{10}} + 2\cos \dfrac{\pi }{2}\cos \dfrac{\pi }{{10}}} \right]\\
= - 2\cos \alpha .\left( {0 + 0} \right)\\
= 0
\end{array}$
