Find whether the function $f(x) = \sin (\sin^{-1}(\sqrt{x-[x]})) + \cos (\sin^{-1}(\sqrt{x-[x]})) - 1$ is even or odd. Where $[x]$ is the G.I.F.
My Attempt:
Firstly, the $\sin x$ part can be simplified cause that's written in $f(f^{-1}(x))$ form.
So the function becomes,
$$f(x) = \sqrt{x-[x]} + \cos (\sin^{-1}(\sqrt{x-[x]})) - 1$$
since $x-[x]$ is the fractional part of $x$ plugging in $-x$ would simply result in $1-(x-[x])$
So, the function becomes,
$$f(-x) = \sqrt{1-x+[x]} + \cos (\sin^{-1}(\sqrt{1-x+[x]})) - 1$$
But I don't know how do I proceed further. My book says that it's an even function.
Any help would be appreciated.
You have found that $$f(-x) = \sqrt{1-x+[x]} + \cos (\sin^{-1}(\sqrt{1-x+[x]})) - 1$$ Note that $$ \cos (\sin^{-1}(\sqrt{1-x+[x]})) = \sqrt{x-[x]}$$
Thus you have $$ f(-x) = \sqrt{1-x+[x]} + \sqrt{x-[x]} -1 = f(x)$$
That is the function is even.