# Find whether the function $\sin (\sin^{-1}(\sqrt{x-[x]})) + \cos (\sin^{-1}(\sqrt{x-[x]})) - 1$ is even or odd?

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.

• $$\cos(\sin^{-1}x)=+\sqrt{1-x^2}$$ – 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 $$f(-x) = \sqrt{1-x+[x]} + \sqrt{x-[x]} -1 = f(x)$$