Question

Sin32° Cos28°-Sin17° Cos13°

Answer 1

$Sin32^{0}Cos28^{0}-Sin17^{0}Cos13^{0}=\frac{Sin(32^{0}-28^{0})+Sin(32^{0}+28^{0})}{2}-\frac{Sin(17^{0}-13^{0})+Sin(17^{0}+13^{0})}{2}=\frac{Sin4^{0}+Sin60^{0}-Sin4^{0}-Sin30^{0}}{2}=\frac{Sin60^{0}-Sin30^{0}}{2}=\frac{\frac{\sqrt{3}}{2}-\frac{1}{2}}{2}=\frac{\sqrt{3}-1}{4}$