Extracting $x$ from $\cos(\arcsin(x))$ The following I know to be valid: $x = \sin(\arcsin(x))$
But is it possible to extract $u$ from $\cos(\arcsin(u))$  ?  
Should it be: $\cos(\arcsin(t)) = \sin\left(\dfrac{\pi}{2} + \arcsin(t)\right) = \dfrac{\pi}{2} + t$ ?
 A: Draw a right triangle such that $\sin\theta = u$ with side lengths $u$, $\sqrt{1-u^2}$ and hypotenuse $1$. Then $$\cos(\arcsin(u)) = \cos\theta = \sqrt{1 - u^2}$$
A: If you're not confident with geometry, you can tackle the problem with analysis; consider the function
$$
f(x)=\cos\arcsin x
$$
which is defined in the interval $[-1,1]$. Its derivative is
$$
f'(x)=-\sin\arcsin x\cdot\frac{1}{\sqrt{1-x^2}}=-\frac{x}{\sqrt{1-x^2}}
$$
for $x\in(-1,1)$. The function $g(x)=\sqrt{1-x^2}$ has derivative
$$
g'(x)=-\frac{x}{\sqrt{1-x^2}}
$$
again for $x\in(-1,1)$. So we know that, for some constant $c$, $f(x)=g(x)+c$ for $x\in(-1,1)$. Since $f(0)=\cos\arcsin0=\cos0=1$ and $g(0)=\sqrt{1-0^2}=1$, we deduce $c=0$. Therefore
$$
\cos\arcsin x=\sqrt{1-x^2}
$$
for $x\in(-1,1)$ and this is valid also for $x=-1$ and $x=1$ by direct check.
A: In general, 
$\sin(x+ \pi /2) = \cos(x),$ so your first equality is correct, in particular when $x=\arcsin(t)$.
For the next equality, \@jef is right let 
$x=\arcsin(u)$.
So we have $\sin(x)=u = \frac{u}{1}=$opposite/hypotenuse.
Drawing the reference triangle, we see the the second leg of the triangle, i.e the adjacent side is $\sqrt{1-u^2}$. Hence, 
$\cos(\arcsin(u))=\cos(x)=$adjacent/hypotenuse=$\sqrt{1-u^2}/1.$
A: Let $\arcsin x=u\implies x=\sin u$ and $-\dfrac\pi2\le u\le\dfrac\pi2\implies\cos u\ge0$
and consequently, $$\cos u=+\sqrt{1-\sin^2u}=+\sqrt{1-x^2}$$
