Show that $\displaystyle \displaystyle \arccos \Big(\frac{\cos x}{1+ 2 \cos x } \Big) = \arctan \sqrt{\frac{1 + \cos x}{1 + 3 \cos x}}$ .
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
I know this is in the trigonometry forum, but I'm sure you won't mind integral calculus!
$\displaystyle \displaystyle \cos^{-1}{x} = \int_{x}^{1}\frac{1}{\sqrt{1-t^2}}\;{dt} = \int_{x}^{1}\frac{1}{\sqrt{1-t}\sqrt{1+t}}\;{dt}$
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} $