How do I simplify:
$\cos(\sin^{-1}\left(\frac {x+1}{\sqrt{2x^2+2}}\right))$
I have tried doing $x=\sin(u)$ and subbing in, but I am getting stuck on what to do after that.
so $\dfrac {\cos((\sin(u)+1)}{\sqrt{2\sin(u)^2+2}}$
is where I'm at, but it doesn't look any better.
On
$$\dfrac1{\sqrt2}\left(\dfrac x{\sqrt{x^2+1}}+\dfrac1{\sqrt{x^2+1}}\right)$$
$$=\sin\left(\dfrac\pi4+\arcsin\dfrac x{\sqrt{x^2+1}}\right)$$
which will be more prominent if we set $y=\arctan x,x=\tan y$ with $-\dfrac\pi2<y<\dfrac\pi2\implies\cos y>0$
Use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
On
$$\cos[\sin^{-1} \frac{x+1}{\sqrt{2x^2+2}}]$$ $$=\sqrt{1-\left(\frac{x+1}{\sqrt{2x^2+2}}\right)^2}=\frac{|x-1|}{\sqrt{2x^2+2}}$$
On
$$\cos\left[\sin^{-1}\frac {x+1}{\sqrt{2x^2+2}}\right]$$
For convenience, let $\theta = \sin^{-1}\dfrac {x+1}{\sqrt{2x^2+2}}$. This means that $\sin \theta = \dfrac {x+1}{\sqrt{2x^2+2}}$.
First keep in mind that the arcsin function takes numbers between $-1$ and $1$ and returns $\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
It follows that $\cos(\theta) \in [0,1]$.
We find
\begin{align} (x+1)^2 + (x-1)^2 &= 2x^2 + 2 \\ \dfrac{(x+1)^2}{2x^2 + 2} + \dfrac{(x-1)^2}{2x^2 + 2} &= 1 \\ \left(\dfrac{x+1}{\sqrt{2x^2 + 2}}\right)^2 + \left(\dfrac{x-1}{\sqrt{2x^2 + 2}}\right)^2 &= 1 \\ \left(\dfrac{x-1}{\sqrt{2x^2 + 2}}\right)^2 &= 1-\left(\dfrac{x+1}{\sqrt{2x^2 + 2}}\right)^2 \\ \hline \left(\cos\left[\sin^{-1}\frac {x+1}{\sqrt{2x^2+2}}\right]\right)^2 &= 1 - \left(\sin\left[\sin^{-1}\frac {x+1}{\sqrt{2x^2+2}}\right]\right)^2 \\ \cos^2(\theta) &= 1 - \sin^2(\theta) \\ \cos^2(\theta) &= \left(\dfrac{x-1}{\sqrt{2x^2 + 2}}\right)^2 \end{align}
Because $\cos(\theta)$ needs to be non negative, we conclude that
$$\cos(\theta) = \dfrac{|x-1|}{\sqrt{2x^2 + 2}}$$
Hint : \begin{eqnarray*} \cos( \sin^{-1}( \theta)) = \sqrt{ 1-(\sin( \sin^{-1}(\theta))^2} =\sqrt{1-\theta^2} . \end{eqnarray*}