# Finding the Big-O of $n^{\sin^2n} \cdot \sqrt{n}$

I need to find the Big-O of $f(n) = n^{\sin^2n} \cdot \sqrt{n}$. I know that the value of $\sin(n)$ oscillates between -1 and 1, and so does the value of $\sin^2(n) = \sin ( \sin(n))$. Now, if I am required to be as accurate as possible, should I say that $f(n) = O(n^{1.5})$ and $f(n) = \Omega(\dfrac{1}{\sqrt{n}})$. Maybe as a more general side question, should I always provide both the $O$ and $\Omega$ notation, if the I cannot express it with $\Theta$?

For me, this feels both correct and accurate. This is not strictly homework, but yes, I am practicing for an exam.

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"$\sin^2 n$" conventionally means $(\sin n)^2$, not $\sin(\sin n)$. – Henning Makholm Sep 18 '11 at 19:41
Indeed you are right and in that case it would be even easier to just say that $f(n) = O(n^{1.5})$. – Pele Sep 18 '11 at 19:49

Your proposition is optimal in the sense that $f(n)=O(n^a)$ is false for $a<3/2$ and $f(n)=\Omega(n^a)$ is false for $a>-1/2$.
Whether you should always provide $O$ and $\Omega$ very much depends on the context.