Cho

- Hướng dẫn giải

Vì $cos\alpha =1$ nên $\alpha =k2\pi$, từ đó

$\sin \left(\alpha -\frac{\mathrm{\pi }}{6}\right)=\sin \left(-\frac{\mathrm{\pi }}{6}\right)=-\frac{1}{2};\sin \left(\alpha +\frac{\mathrm{\pi }}{6}\right)=\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}$.

${\sin }^2\left(\alpha -\frac{\mathrm{\pi }}{6}\right)+{\sin }^2\left(\alpha +\frac{\mathrm{\pi }}{6}\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$