# Give upper and lower bound for $\left|\cos{na} + \frac{\sin{nb}}{n}\right|$

The question asks to give the upper and lower estimates (in terms of $n$ alone), for $$\left|\cos{na} + \frac{\sin{nb}}{n}\right|$$ valid for all $a, b$. Show these estimates are the best possible by giving values $a$ and $b$ for which the bounds are actually attained.

I found the upper bound to be $1 + 1/n$, which wasn't too hard, but I'm not sure how to find the lower bound, and I don't understand what the second half of the question is asking for.

Update:

If I apply the reverse inequality, should the lower bound be $0$ or $1/n$?

• Take $a=\pi/\left(2n\right),\, b=0$ and then your absolute value is $0$. This is your lower bound. Dec 1, 2014 at 12:55

Let us ignore for the moment the case of $n=0$.
The lower bound of both the $\cos$ and $\sin$ terms are $0$, independently. Take $a=\pi/2n$ and $b=\pi/n$ (other $(a,b)$ combinations can do). So, $0$.
The upper bound of the $\cos$ term is $1$, and that of the $\sin$ terms is $1/n$, independently. Take $a=0$ and $b=\pi/2n$. So, $1+1/n$.
Now for $n=0$, either you consider that the $\sin$ term is undefined, of you take it equal to its limit $n\to0$, i.e. $b$, and you need to bound $|1+b|$. The lower bound is $1$ and there is no upper bound.