# identity

• Feb 19th 2011, 04:37 PM
Random Variable
identity
Show that $\displaystyle \arccos \Big(\frac{\cos x}{1+ 2 \cos x } \Big) = \arctan \sqrt{\frac{1 + \cos x}{1 + 3 \cos x}}$ .
• Feb 19th 2011, 05:31 PM
Ackbeet
This identity is not true, I don't think. Let $x=\pi/4.$ Then the LHS is approximately $1.27,$ and the RHS is approximately $0.64.$
• Feb 19th 2011, 06:35 PM
Random Variable
My mistake. It should be $\displaystyle \arccos \Big(\frac{\cos x}{1+ 2 \cos x } \Big) = 2 \arctan \sqrt{\frac{1 + \cos x}{1 + 3 \cos x}}$
• Feb 26th 2011, 07:04 PM
Random Variable
The identity being used is $\displaystyle \arccos x = 2 \arctan \sqrt{\frac{1-x}{1+x}}$. But from where does that identity come?
• Feb 26th 2011, 07:27 PM
topsquark
Quote:

Originally Posted by Random Variable
The identity being used is $\displaystyle \arccos x = 2 \arctan \sqrt{\frac{1-x}{1+x}}$. But from where does that identity come?

Hang on. This is going to be a ride...

Take the cosine of both sides:
$\displaystyle cos(acs x) = x$

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

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

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

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

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

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

Square both of these and subtract and you get
$\displaystyle cos^2( \theta ) - sin^2( \theta ) = \frac{1 + x}{2} - \frac{1 - x}{2} = x$

-Dan
• Feb 26th 2011, 08:35 PM
TheCoffeeMachine
Quote:

Originally Posted by Random Variable
The identity being used is $\displaystyle \arccos x = 2 \arctan \sqrt{\frac{1-x}{1+x}}$. But from where does that identity come?

I know this is in the trigonometry forum, but I'm sure you won't mind integral calculus! :p

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

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

\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 PM
sa-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.