justinelee127
justinelee127
02-06-2017
Mathematics
contestada
rewrite 2tan3x in terms of tanx
Respuesta :
LammettHash
LammettHash
02-06-2017
[tex]2\tan3x=2\tan(2x+x)=\dfrac{2\tan2x+2\tan x}{1-\tan2x\tan x}[/tex]
Use the same identity to expand [tex]\tan2x[/tex].
[tex]\tan2x=\tan(x+x)=\dfrac{\tan x+\tan x}{1-\tan x\tan x}=\dfrac{2\tan x}{1-\tan^2x}[/tex]
[tex]\implies2\tan3x=\dfrac{2\dfrac{2\tan x}{1-\tan^2x}+2\tan x}{1-\dfrac{2\tan x}{1-\tan^2x}\tan x}[/tex]
[tex]2\tan3x=\dfrac{2\tan x\left(\dfrac2{1-\tan^2x}+1\right)}{1-\dfrac{2\tan^2x}{1-\tan^2x}}[/tex]
[tex]2\tan3x=\dfrac{2\tan x\left(2+1-\tan^2x\right)}{1-\tan^2x-2\tan^2x}[/tex]
[tex]2\tan3x=\dfrac{2\tan x(3-\tan^2x)}{1-3\tan^2x}[/tex]
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