Question Number 200040 by mnjuly1970 last updated on 12/Nov/23
$$ \\ $$$$\:\:\:\:\:{Q}:\:\:{If}\:\:,\:\:{tan}\left(\frac{\pi}{\mathrm{4}}\:−\alpha\:\right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\Rightarrow{Find}\:{the}\:{value}\:{of}\:,\:{tan}\left(\mathrm{4}\alpha\right)=? \\ $$$$ \\ $$
Answered by witcher3 last updated on 12/Nov/23
$$\Rightarrow\mathrm{tg}\left(\pi−\mathrm{4}\alpha\right)=−\mathrm{tg}\left(\mathrm{4}\alpha\right) \\ $$$$\mathrm{tg}\left(\mathrm{4x}\right).\frac{\mathrm{2tg}\left(\mathrm{2x}\right)}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \left(\mathrm{2x}\right)}=\frac{\mathrm{2}.\frac{\mathrm{2tg}\left(\mathrm{x}\right)}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)}}{\mathrm{1}−\left(\frac{\mathrm{2tg}\left(\mathrm{x}\right)}{\mathrm{1}−\mathrm{tg}^{\mathrm{2}} \left(\mathrm{x}\right)}\right)^{\mathrm{2}} } \\ $$$$\mathrm{tg}\left(\mathrm{4}\left(\frac{\pi}{\mathrm{4}}−\alpha\right)=\frac{\mathrm{2}\frac{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{1}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} }=\frac{\frac{\mathrm{8}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{16}}{\mathrm{9}}}=−\frac{\mathrm{24}}{\mathrm{7}}=−\mathrm{tg}\left(\mathrm{4a}\right)\right. \\ $$$$\mathrm{tg}\left(\mathrm{4a}\right)=\frac{\mathrm{24}}{\mathrm{7}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 13/Nov/23
$$\:\:\:{thx}\:\:\:{alot}… \\ $$
Answered by des_ last updated on 12/Nov/23
$$\frac{\mathrm{1}−\mathrm{tan}\left(\alpha\right)}{\mathrm{1}+\mathrm{tan}\left(\alpha\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\mathrm{tan}\left(\alpha\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:; \\ $$$$\mathrm{tan}\left(\mathrm{2}\alpha\right)\:=\:\frac{\mathrm{2tan}\left(\alpha\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left(\alpha\right)}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:; \\ $$$$\mathrm{tan}\left(\mathrm{4}\alpha\right)\:=\:\frac{\mathrm{2tan}\left(\mathrm{2}\alpha\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left(\mathrm{2}\alpha\right)}\:=\:\frac{\mathrm{24}}{\mathrm{7}} \\ $$
Commented by mnjuly1970 last updated on 13/Nov/23
$${thx}\:{alot} \\ $$