Question Number 226780 by Spillover last updated on 14/Dec/25
$${By}\:{using}\:{concept}\:{of}\:{complex} \\ $$$${number} \\ $$$${show}\:{that} \\ $$$$\mathrm{tan}\:\mathrm{5}\theta=\frac{\mathrm{tan}\:^{\mathrm{5}} \theta−\mathrm{10tan}\:^{\mathrm{3}} \theta+\mathrm{5tan}\:\theta}{\mathrm{5tan}\:^{\mathrm{4}} \theta−\mathrm{10tan}\:^{\mathrm{2}} \theta+\mathrm{1}} \\ $$
Answered by Frix last updated on 14/Dec/25
$$\mathrm{tan}\:\theta\:=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{2i}\theta} }{\mathrm{1}+\mathrm{e}^{\mathrm{2i}\theta} }\mathrm{i}={t} \\ $$$$\frac{\mathrm{t}^{\mathrm{5}} −\mathrm{10}{t}^{\mathrm{3}} +\mathrm{5}{t}}{\mathrm{5}{t}^{\mathrm{4}} −\mathrm{10}{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{10i}\theta} }{\mathrm{1}+\mathrm{e}^{\mathrm{10i}\theta} }\mathrm{i}=\frac{\mathrm{1}−\mathrm{e}^{\mathrm{2i}\left(\mathrm{5}\theta\right)} }{\mathrm{1}+\mathrm{e}^{\mathrm{2i}\left(\mathrm{5}\theta\right)} }\mathrm{i}=\mathrm{tan}\:\mathrm{5}\theta \\ $$