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tan(2 cos^-1(x))

Elissa89

Member
Oct 19, 2017
52
tan(2 cos^-1(x))

I'm pretty sure 2 cos^-1(x) is the same as cos2x, the doubles formula, but from there I'm completely lost.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
644
tan(2 cos^-1(x))

I'm pretty sure 2 cos^-1(x) is the same as cos2x, the doubles formula, but from there I'm completely lost.
$2\cos^{-1}(x) \ne \cos(2x)$

$\cos^{-1}(x)$ is an angle (call it $\theta$) such $\cos{\theta} = x$

$\tan(2\theta) = \dfrac{2\tan{\theta}}{1-\tan^2{\theta}}$

$\cos{\theta} = x \implies \sin{\theta} = \sqrt{1-x^2} \implies \tan{\theta} = \dfrac{\sqrt{1-x^2}}{x}$

substitute the above expression for $\tan{\theta}$ in the double angle formula for tangent and simplify the algebra.
 
Last edited:

Elissa89

Member
Oct 19, 2017
52
$2\cos^{-1}(x) \ne \cos(2x)$

$\cos^{-1}(x)$ is an angle (call it $\theta$) such $\cos{\theta} = x$

$\tan(2\theta) = \dfrac{2\tan{\theta}}{1-\tan^2{\theta}}$

$\cos{\theta} = x \implies \sin{\theta} = \sqrt{1-x^2} \implies \tan{\theta} = \dfrac{\sqrt{1-x^2}}{x}$

substitute the above expression for $\tan{\theta}$ in the double angle formula for tangent and simplify the algebra.
I'm not sure I understand
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,120

DavidCampen

Member
Apr 4, 2014
64
We are having to guess at what the OP is supposed to accomplish. Is it calculating a value for the expression given a value for x or, as we are guessing, is it to find a purely algebraic expression for the given trigonometric expression; or even something else.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
376
It appears that you do not understand what "[tex]cos^{-1}(x)[/tex]" means! It is the inverse function to cos(x) (limited to x= 0 to [tex]\pi[/tex] so that it is one-to-one). [tex]cos^{-1}(1)= 0[/tex] because cos(0)= 1. [tex]cos^{-1}(0)= \pi/2[/tex] because [tex]cos(\pi/2)= 0[/tex], etc.

The basic fact you need to use here is that [tex]cos(cos^{-1}(x))= x[/tex] so you should try to change that "tan(2..)" to "cos(..)". I would start by using the identity [tex]tan(2x)= \frac{2 tan(x)}{1- tan^2(x)}[/tex]. Now use the fact that [tex]tan(x)= \frac{sin(x)}{cos(x)}[/tex] and then that [tex]sin(x)= \sqrt{1- cos^2(x)}[/tex]: [tex]tan(2x)= \frac{2 tan(x)}{1- tan^2(x)}[/tex][tex]= \frac{2\frac{\sqrt{1- cos^2(x)}}{cos(x)}}{1- \frac{1- cos^2(x)}{cos^2(x)}}[/tex].

Replacing each "x" with [tex]cos^{-1}(x)[/tex] changes each "cos(x)" to "x" so that
[tex]tan(2cos^{-1}(x))= \frac{2\frac{\sqrt{1- x^2}}{x}}{1- \frac{1- x^2}{x^2}}= \frac{2x\sqrt{1- x^2}}{2x^2- 1}[/tex].