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Question Number 76497 by mathmax by abdo last updated on 27/Dec/19

calculate tan(3x) interms of tanx

$${calculate}\:{tan}\left(\mathrm{3}{x}\right)\:{interms}\:{of}\:{tanx} \\ $$

Answered by Rio Michael last updated on 27/Dec/19

tan3x = tan(2x + x) = ((tan2x + tanx)/(1−tan2xtanx)) = ((((2tanx)/(1−tan^2 x)) + tanx)/(1− ((2tanx)/(1−tan^2 x)) tanx)) = ((2tanx+ tanx − tan^3 x)/(1−tan^2 x−2tan^3 x)) = ((3tanx−tan^3 x)/(1−tan^2 x−2tan^3 x))

$${tan}\mathrm{3}{x}\:=\:{tan}\left(\mathrm{2}{x}\:+\:{x}\right)\:=\:\frac{{tan}\mathrm{2}{x}\:+\:{tanx}}{\mathrm{1}−{tan}\mathrm{2}{xtanx}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}\:+\:{tanx}}{\mathrm{1}−\:\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}\:{tanx}}\:=\:\frac{\mathrm{2}{tanx}+\:{tanx}\:−\:{tan}^{\mathrm{3}} {x}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}−\mathrm{2}{tan}^{\mathrm{3}} {x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3}{tanx}−{tan}^{\mathrm{3}} {x}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}−\mathrm{2}{tan}^{\mathrm{3}} {x}} \\ $$

Commented by mathmax by abdo last updated on 27/Dec/19

thankx sir.

$${thankx}\:{sir}. \\ $$

Commented by MJS last updated on 27/Dec/19

=((tan x (tan^2 x −3))/(3tan^2 x −1)) error in line 2

$$=\frac{\mathrm{tan}\:{x}\:\left(\mathrm{tan}^{\mathrm{2}} \:{x}\:−\mathrm{3}\right)}{\mathrm{3tan}^{\mathrm{2}} \:{x}\:−\mathrm{1}} \\ $$$$\mathrm{error}\:\mathrm{in}\:\mathrm{line}\:\mathrm{2} \\ $$

Answered by $@ty@m123 last updated on 28/Dec/19

tan 3x=((sin 3x)/(cos 3x)) =((3sin x−4sin^3 x)/(4cos^3 x−3cos x)) =(((3sin x−4sin^3 x)÷cos^3 x)/((4cos^3 x−3cos x)÷cos^3 x)) =((3tan x.sec^2 x−4tan^3 x)/(4−3sec^2 x)) =((3tan x.(1+tan^2 x)−4tan^3 x)/(4−3(1+tan^2 x))) =((3tan x+3tan^3 x−4tan^3 x)/(4−3−3tan^2 x)) =((3tan x−tan^3 x)/(1−3tan^2 x))

$$\mathrm{tan}\:\mathrm{3}{x}=\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{3}{x}} \\ $$$$=\frac{\mathrm{3sin}\:{x}−\mathrm{4sin}\:^{\mathrm{3}} {x}}{\mathrm{4cos}\:^{\mathrm{3}} {x}−\mathrm{3cos}\:{x}} \\ $$$$=\frac{\left(\mathrm{3sin}\:{x}−\mathrm{4sin}\:^{\mathrm{3}} {x}\right)\boldsymbol{\div}\mathrm{cos}\:^{\mathrm{3}} {x}}{\left(\mathrm{4cos}\:^{\mathrm{3}} {x}−\mathrm{3cos}\:{x}\right)\boldsymbol{\div}\mathrm{cos}\:^{\mathrm{3}} {x}} \\ $$$$=\frac{\mathrm{3tan}\:{x}.\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{4tan}\:^{\mathrm{3}} {x}}{\mathrm{4}−\mathrm{3sec}\:^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{3tan}\:{x}.\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)−\mathrm{4tan}\:^{\mathrm{3}} {x}}{\mathrm{4}−\mathrm{3}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}\right)} \\ $$$$=\frac{\mathrm{3tan}\:{x}+\mathrm{3tan}\:^{\mathrm{3}} {x}−\mathrm{4tan}\:^{\mathrm{3}} {x}}{\mathrm{4}−\mathrm{3}−\mathrm{3tan}\:^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{3tan}\:{x}−\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} {x}} \\ $$

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