Question Number 17119 by gourav~ last updated on 01/Jul/17
Commented by prakash jain last updated on 01/Jul/17
$$\frac{\mathrm{sin}\:\left({A}+\mathrm{3}{B}\right)+\mathrm{sin}\:\left(\mathrm{3}{A}+{B}\right)}{\mathrm{sin}\:\mathrm{2}{A}+\mathrm{sin}\:\mathrm{2}{B}} \\ $$$$=\frac{\mathrm{2sin}\:\left(\frac{\mathrm{4}{A}+\mathrm{4}{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}{B}−\mathrm{2}{A}}{\mathrm{2}}\right)}{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)} \\ $$$$=\frac{\mathrm{2sin}\:\left(\mathrm{2}\left({A}+{B}\right)\right)\mathrm{cos}\:\left({A}−{B}\right)}{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)} \\ $$$$=\frac{\mathrm{sin}\:\left(\mathrm{2}\left({A}+{B}\right)\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\ $$$$=\frac{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}+{B}\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\ $$$$=\mathrm{2cos}\:\left({A}+{B}\right)\blacksquare \\ $$
Commented by gourav~ last updated on 01/Jul/17
$${thank}\:{you}\:{sir} \\ $$