MHB Understanding $\sin\left({\frac{1}{n^2}}\right)$ When $n > 1$

[tmt1]

Given than $n > 1$, then $\sin\left({\frac{1}{n^2}}\right) > 0$, but I'm not sure why that is.

I get that a sin function in the first quadrant will yield a positive result, but I'm not sure why it's in the first quadrant in the first place. Would $\frac{1}{n^2}$ be in degrees in this case, in which case, since it's less than 1 it would be in the first quadrant. That makes sense, but I'm not sure how to know that $n$ would be notated in degrees.

 

[Greg]

$0\lt\dfrac{1}{n^2}\lt\dfrac{\pi}{2}$ hence $0\lt\sin\left(\dfrac{1}{n^2}\right)\lt1$ for $n>1$.

 

[HallsofIvy]

If n> 1 then n^2> 1 so \frac{1}{n^2}< 1 so \frac{1}{n^2} is certainly less than either 90 (degrees) or \frac{\pi}{2} radians.