Question Number 52629 by scientist last updated on 10/Jan/19
$${show}\:{that}\:\frac{{sin}\alpha+{sin}\mathrm{3}\alpha+{sin}\mathrm{5}\alpha}{{cos}\alpha+{cos}\mathrm{3}\alpha+{cos}\mathrm{5}\alpha}=\:{tan}\mathrm{3}\alpha \\ $$
Answered by math1967 last updated on 10/Jan/19
$$\frac{{sin}\mathrm{3}\alpha+\mathrm{2sin}\:\mathrm{3}\alpha.\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{cos}\:\mathrm{3}\alpha+\mathrm{2cos}\:\mathrm{3}\alpha\mathrm{cos}\:\mathrm{2}\alpha} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{3}\alpha\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right)}{\mathrm{cos}\:\mathrm{3}\alpha\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right)}={tan}\mathrm{3}\alpha \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19
$$\frac{{sin}\mathrm{5}\alpha+{sin}\alpha+{sin}\mathrm{3}\alpha}{{cos}\mathrm{5}\alpha+{cos}\alpha+{cos}\mathrm{3}\alpha} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{3}\alpha.{cos}\mathrm{2}\alpha+{sin}\mathrm{3}\alpha}{\mathrm{2}{cos}\mathrm{3}\alpha{cos}\mathrm{2}\alpha+{cos}\mathrm{3}\alpha} \\ $$$$=\frac{{sin}\mathrm{3}\alpha\left(\mathrm{2}{cos}\mathrm{2}\alpha\:+\mathrm{1}\right)}{{cos}\mathrm{3}\alpha\left(\mathrm{2}{cos}\mathrm{2}\alpha\:+\mathrm{1}\right)} \\ $$$$={tan}\mathrm{3}\alpha \\ $$