Consider the set $E=\{cos \frac{n\pi}{3}+\sin \frac{n\pi}{3}:n\in \mathbb{N}\}$. Then supremum of E and infimum of E is
(a) 1 and -1
(b) $\frac{\sqrt{3}}{2}+\frac{1}{2}$ and $\frac{1}{2}-\frac{\sqrt{3}}{2}$
(c)$\frac{\sqrt{3}}{2}+\frac{1}{2}$ and $-\frac{1}{2}-\frac{\sqrt{3}}{2}$
(d)$\frac{\sqrt{3}}{2}+1$ and $-1-\frac{\sqrt{3}}{2}$
I understood the sum but how to deal with the sum, I am not getting or what is the correct way to deal with the sum...Taking odd even case was bit difficult for me to conclude.. so i started with case where $n$ is even multiple of 3, $n= 2k$, where $k$ is even, in which case $cos \frac{n\pi}{3}+\sin \frac{n\pi}{3}=-1$.
Then for the case where $n=3(2k+1)$, here $cos \frac{n\pi}{3}+\sin \frac{n\pi}{3}=1$
But this this not the way to deal with this.
I can see, similarly terms in option (b) and (c) also belong to the set $E$. But how to get sure there cannot be any elements bigger than $\frac{\sqrt{3}}{2}+\frac{1}{2}$ and smaller than $\frac{-\sqrt{3}}{2}-\frac{1}{2}$ in the set to conclude it as the supremum and infimum respectively
But elements of $d$ will not belong to the set $E$, and obviously supremum and infimum need not belong to the set, so they can still be supremum or infimum.
Or is there a way to deal with this type problem in a better way.
thanks in advnce!!
Just check the possible values of both trig. functions for different $\;n\;$ . They're the similar but at different points, of course:
$$\cos\frac{n\pi}3=\begin{cases}\frac12\;,\;\;n=\pm1\pmod 6\\{}\\ -\frac12\,,\;\;n=2,4\pmod 6\\{}\\ -1\,,\;\;n=3\pmod 6\\{}\\ 1\,,\;\;n=0\pmod 6\end{cases}$$
Of course, the above can be made even easier if you think of the positive/negative values of cosine and sine in the the different quadrants.
Now you do the same as above for the sine function and it then be clear what are the maximum/minimum values of the sum.