Q:

Solving Inverse trig problems using substitution?

I have this problem
$$\arccos\left(\frac{x+\sqrt{1-x^2}}{\sqrt{2}}\right)$$
The answer comes out be $\arcsin(x)-\frac{\pi}{4}$
I've realized that this problem can be solved by using something called substitution, but i really dont get the idea of how you can just substitute $x$ with $\cos(x),~\sin(x)$. Or anything else for that matter.
Also how do you know what to substitute? Is there a method for that?
This has been confusing me a lot and i would appreciate if the answer is not just the solution but also an explanation to how substitution works in brief.
Thanks in advance.

A:

I assume you want to simplify the expression.

$\sqrt{1-x^2}$ is defined only when $|x|\leq 1$. Hence, if we let $x=\sin \alpha$, note that for every possible value of $x$ we can select a value of $\alpha$.
Let $t=\arccos\left( \dfrac{x+\sqrt{1-x^2}}{\sqrt 2}\right)$.
Now,
$$t=\arccos \left(\dfrac{\sin \alpha + \cos \alpha}{\sqrt 2}\right)$$
Or
$$t=\arccos \left(\cos (\alpha-\pi/4)\right)=\alpha -\pi/4$$
This gives us $t=\arcsin x -\pi/4$.
Note: the above is valid only for certain values of $\alpha$. I have left this for you as an exercise(the values for which it is valid).
Edit:
$  t = \begin{cases}
\arcsin x-\pi/4,  & 1\geq x\geq 1/\sqrt 2 \\
\pi/4 -\arcsin x, & 1/\sqrt 2\geq x\geq -1
\end{cases}$
Corresponding Desmos plot:

