Question

sin(23π/24))^3cos(π/24)+(cos(23π/24))^3sin(π/24)

Answer 1

$\displaystyle sin^3(\frac{23\pi}{24})cos(\frac{\pi}{24})+cos^3(\frac{23\pi}{24})sin(\frac{\pi}{24})=\\=sin^3(\pi-\frac{\pi}{24})cos(\frac{\pi}{24})+cos^3(\pi-\frac{\pi}{24})sin(\frac{\pi}{24})=\\=sin^3(\frac{\pi}{24})cos(\frac{\pi}{24})-cos^3(\frac{\pi}{24})sin(\frac{\pi}{24})=\\=sin(\frac{\pi}{24})cos(\frac{\pi}{24})(sin^2(\frac{\pi}{24})-cos^2(\frac{\pi}{24}))=-\frac{1}{2}sin(\frac{\pi}{12})cos(\frac{\pi}{12})=\\=-\frac{1}{4}sin(\frac{\pi}{6})=-\frac{1}{8}$