Question Number 42520 by gyugfeet last updated on 27/Aug/18
$${cos}^{\mathrm{3}} \:{A}.{sin}\mathrm{3}{A}+{sinA}.{cos}\mathrm{3}{A}=\frac{\mathrm{3}}{\mathrm{4}}{sin}\mathrm{4}{A} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18
$${cos}\mathrm{3}{A}=\frac{\mathrm{4}{cos}^{\mathrm{3}} {A}−\mathrm{3}{cosA}}{} \\ $$$$\left(\frac{\mathrm{3}{cosA}+{cos}\mathrm{3}{A}}{\mathrm{4}}\right){sin}\mathrm{3}{A}+{sinAcos}\mathrm{3}{A} \\ $$$$\frac{\mathrm{3}{cosAsin}\mathrm{3}{A}+{cos}\mathrm{3}{Asin}\mathrm{3}{A}+\mathrm{4}{sinAcos}\mathrm{3}{A}}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}\left({cosAsin}\mathrm{3}{A}+{sinAcos}\mathrm{3}{A}\right)+{cos}\mathrm{3}{Asin}\mathrm{3}{A}+{sinAcos}\mathrm{3}{A}}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}{sin}\mathrm{4}{A}+{cos}\mathrm{3}{A}\left({sin}\mathrm{3}{A}+{sinA}\right)}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}{sin}\mathrm{4}{A}+\frac{\mathrm{1}}{\mathrm{4}}{cos}\mathrm{3}{A}\left(\mathrm{2}{sin}\mathrm{2}{AcosA}\right) \\ $$$${pls}\:{check}\:{the}\:{auestion}… \\ $$