Cho cos anpha/2 = 0. Khi đó cos anpha + cos 2anpha + cos 4anpha + cos7anpha bằng

* Hướng dẫn giải

$$\begin{gathered} \cos \alpha =2.{\cos }^2\frac{\alpha }{2}-1=2.0-1=-1 \\ \cos 2\alpha =2{\cos }^2\alpha -1=2.{(-1)}^2-1=1 \\ \cos 4\alpha =2{\cos }^{2}2\alpha -1=2.1^2-1=1 \\ \cos 7\alpha =\cos \left(a+6a\right)=\cos \alpha .\cos 6\alpha -\sin \alpha \sin 6\alpha =-\cos (3.2\alpha )=-4{\cos }^{3}2\alpha +3\cos 2\alpha =-4+3=-1 \\ \cos \alpha +\cos 2\alpha +\cos 4\alpha +\cos 7\alpha =-1+1+1+(-1)=0 \end{gathered}$$