Question Number 164757 by mathlove last updated on 21/Jan/22
$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)−\mathrm{3}{sin}^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{2}}+\alpha\right)−\mathrm{2}{cos}\left(\mathrm{5}\pi+\alpha\right)=? \\ $$
Commented by blackmamba last updated on 21/Jan/22
$$\Rightarrow=\mathrm{cos}\:\alpha−\mathrm{3cos}\:^{\mathrm{2}} \alpha+\mathrm{2cos}\:\alpha \\ $$$$\Rightarrow=\mathrm{3}×\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{3}×\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\Rightarrow=\frac{\mathrm{6}−\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$