Question Number 6146 by Ninik last updated on 16/Jun/16
$${Determine}\:{the}\:{value}\:{of}\:\mathrm{sin}\:\left(\frac{\Pi}{\mathrm{3}}+{p}\right)\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{6}}+{p}\right)−\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{3}}+{p}\right)\mathrm{sin}\:\left(\frac{\Pi}{\mathrm{6}}+{p}\right) \\ $$
Answered by Rasheed Soomro last updated on 16/Jun/16
$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+{p}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+{p}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+{p}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+{p}\right) \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$=\mathrm{sin}\:\left(\left(\frac{\pi}{\mathrm{3}}+{p}\right)−\left(\frac{\pi}{\mathrm{6}}+{p}\right)\right) \\ $$$$=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{6}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$