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.

  • 2
    $\begingroup$ $$\cos(\sin^{-1}x)=+\sqrt{1-x^2}$$ $\endgroup$ – lab bhattacharjee Jun 20 '19 at 14:43

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.


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