Question Number 10917 by okhema last updated on 02/Mar/17
$${hence}\:{show}\:{that} \\ $$$$\left.{i}\right)\frac{\mathrm{1}−{cos}\mathrm{4}\theta}{{sin}\mathrm{4}\theta}={tan}\mathrm{2}\theta \\ $$$$\left.{ii}\right)\frac{\mathrm{1}−{cos}\mathrm{6}\theta}{{sin}\mathrm{6}\theta}={tan}\mathrm{3}\theta \\ $$
Answered by sandy_suhendra last updated on 02/Mar/17
$$\mathrm{we}\:\mathrm{use}\:\:\:\:\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta\:\Rightarrow\:\mathrm{2sin}^{\mathrm{2}} \theta=\mathrm{1}−\mathrm{cos2}\theta \\ $$$$\mathrm{and}\:\:\:\mathrm{sin2}\theta=\mathrm{2sin}\theta\mathrm{cos}\theta \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{same}\:\mathrm{way}\:\:\mathrm{1}−\mathrm{cos4}\theta=\mathrm{2sin}^{\mathrm{2}} \mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−\mathrm{cos6}\theta=\mathrm{2sin}^{\mathrm{2}} \mathrm{3}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sin4}\theta=\mathrm{2sin2}\theta\mathrm{cos2}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sin6}\theta=\mathrm{2sin3}\theta\mathrm{cos3}\theta \\ $$$$\left.\mathrm{i}\right)\:\frac{\mathrm{2sin}^{\mathrm{2}} \mathrm{2}\theta}{\mathrm{2sin2}\theta\mathrm{cos2}\theta}\:=\:\frac{\mathrm{sin2}\theta}{\mathrm{cos2}\theta}\:=\:\mathrm{tan2}\theta \\ $$$$\left.\mathrm{ii}\right)\:\frac{\mathrm{2sin}^{\mathrm{2}} \mathrm{3}\theta}{\mathrm{2sin3}\theta\mathrm{cos3}\theta}\:=\:\frac{\mathrm{sin3}\theta}{\mathrm{cos3}\theta}\:=\:\mathrm{tan3}\theta \\ $$