Is Tan(cos2x) equal to tan(2 cos^-1(x))?

[Elissa89]

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]

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.

 

[Elissa89]

$2\cos^{-1}(x) \ne \cos(2x)$

$\cos^{-1}(x)$ 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]

I'm not sure I understand

What, specifically, don't you understand? We can help you better if we know this.

-Dan

 

[DavidCampen]

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.

 

[HOI]

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].