Show that $\displaystyle \displaystyle \arccos \Big(\frac{\cos x}{1+ 2 \cos x } \Big) = \arctan \sqrt{\frac{1 + \cos x}{1 + 3 \cos x}}$ .

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- Feb 19th 2011, 04:37 PMRandom Variableidentity
Show that $\displaystyle \displaystyle \arccos \Big(\frac{\cos x}{1+ 2 \cos x } \Big) = \arctan \sqrt{\frac{1 + \cos x}{1 + 3 \cos x}}$ .

- Feb 19th 2011, 05:31 PMAckbeet
This identity is not true, I don't think. Let $\displaystyle x=\pi/4.$ Then the LHS is approximately $\displaystyle 1.27,$ and the RHS is approximately $\displaystyle 0.64.$

- Feb 19th 2011, 06:35 PMRandom Variable
My mistake. It should be $\displaystyle \displaystyle \arccos \Big(\frac{\cos x}{1+ 2 \cos x } \Big) = 2 \arctan \sqrt{\frac{1 + \cos x}{1 + 3 \cos x}}$

- Feb 26th 2011, 07:04 PMRandom Variable
The identity being used is $\displaystyle \displaystyle \arccos x = 2 \arctan \sqrt{\frac{1-x}{1+x}} $. But from where does that identity come?

- Feb 26th 2011, 07:27 PMtopsquark
Hang on. This is going to be a ride...

Take the cosine of both sides:

$\displaystyle \displaystyle cos(acs x) = x$

So

$\displaystyle \displaystyle cos \left ( 2 ~ atn \left ( \sqrt{\frac{1-x}{1+x}} \right ) \right )$ must be equal to x.

To check this. let $\displaystyle \theta = atn \left ( \sqrt{\frac{1-x}{1+x}} \right ) \right )$

So we have

$\displaystyle \displaystyle cos \left ( 2 ~ atn \left ( \sqrt{\frac{1-x}{1+x}} \right ) \right ) = cos( 2 \theta) = cos^2(\theta) - sin^2(\theta) $

Now sketch a right triangle. Label one angle $\displaystyle \theta$. The leg of the triangle across from $\displaystyle \theta$ has a length of $\displaystyle \sqrt{1 - x}$ and the other leg is $\displaystyle \sqrt{1 + x}$. This means the hypotenuse has a length of $\displaystyle \sqrt{2}$.

So

$\displaystyle \displaystyle cos( \theta ) = \frac{\sqrt{1 + x}}{\sqrt{2}}$

and

$\displaystyle \displaystyle sin( \theta ) = \frac{\sqrt{1 - x}}{\sqrt{2}}$

Square both of these and subtract and you get

$\displaystyle \displaystyle cos^2( \theta ) - sin^2( \theta ) = \frac{1 + x}{2} - \frac{1 - x}{2} = x$

as advertised.

-Dan - Feb 26th 2011, 08:35 PMTheCoffeeMachine
I know this is in the trigonometry forum, but I'm sure you won't mind integral calculus! :p

$\displaystyle \displaystyle \cos^{-1}{x} = \int_{x}^{1}\frac{1}{\sqrt{1-t^2}}\;{dt} = \int_{x}^{1}\frac{1}{\sqrt{1-t}\sqrt{1+t}}\;{dt}$

Let $\displaystyle u = \frac{\sqrt{1-t}}{\sqrt{1+t}} \Rightarrow dt = -\sqrt{1-t}\sqrt{1+t}(1+t)}\;{du}$ and (for simplicity) write $\displaystyle \alpha = \frac{\sqrt{1-x}}{\sqrt{1+x}}$, then:

$\displaystyle \displaystyle \begin{aligned} \cos^{-1}{x} & = -\int_{\alpha}^{0}\frac{\sqrt{1-t}\sqrt{1+t}(1+t)}{\sqrt{1-t}\sqrt{1+t}}\;{du} = \int_{0}^{\alpha} 1+t \;{du} = \int_{0}^{\alpha} 1+\frac{1-u^2}{1+u^2}\;{du} \\& = \int_{0}^{\alpha} \frac{(1+u^2)+(1-u^2)}{1+u^2}\;{du} = \int_{0}^{\alpha} \frac{2}{1+u^2}\;{du} = 2\tan^{-1}{u}\bigg|_{0}^{\alpha} \\& = 2\tan^{-1}{\alpha}-2\tan^{-1}(0) = 2\tan^{-1}{\alpha} = 2\tan^{-1}\left(\frac{\sqrt{1-x}}{\sqrt{1+x}}\right).\end{aligned} $ - Feb 26th 2011, 09:01 PMsa-ri-ga-ma
Try this method.

Let x = cosθ.

Then sqrt[{1-x}/{1+x}] = sqrt[{1-cosθ}/{1+cosθ)] = sqrt[{2sin^2(θ/2)}/{2cos^2(θ/2)] = tan(θ/2)

Rest is obvious.