- #1

Elissa89

- 52

- 0

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

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- Thread starter Elissa89
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- #1

Elissa89

- 52

- 0

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

- #2

skeeter

- 1,104

- 1

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.

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- #3

Elissa89

- 52

- 0

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

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

- #4

- 1,804

- 740

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

-Dan

- #5

DavidCampen

- 65

- 0

- #6

HOI

- 923

- 2

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].

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