Menu Close

Q-If-tan-pi-4-1-2-Find-the-value-of-tan-4-




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

Leave a Reply

Your email address will not be published. Required fields are marked *