<s>[INST] Which are the $\max$ and $\min$'s of $\frac{\sin{2t}}{2}$, $t \in [0, 2\pi]$? I have to find the max and mins of:
$f(x,y) = xy $ in set $ x²+y²<=1$
After derivating and finding $(0,0)$ as saddle point by hessian...
and testing the border:
$x = cost$, $y = sentf(cost,sent) = 1/2.sin(2t)..$
Wich are the max and mins of $(1/2)*sin(2t)$, $0<=t<=2pi$?
I know that critical points are $t = \frac{\pi}{4}$ and $t = \frac{3\pi}{4}$
But the book shows $t = \frac{5\pi}{4}$ and $t = \frac{7\pi}{4}$ critical points too.Why? Someone can help?
Thanks!
 [/INST] As indicated in Maximilian Janisch's question comment, since $\sin(x)$ is periodic with period $2\pi$, you have that $\sin(2t)$ is periodic with period $\frac{2\pi}{2} = \pi$.
Thus, the values of $\sin(2t)$ for $t = \frac{\pi}{4}$ and $t = \frac{5\pi}{4} = \frac{\pi}{4} + \pi$ are the same, with both giving $\sin(2t) = 1$, so $\frac{\sin(2t)}{2} = \frac{1}{2}$ is a maximum at both points. Also, those for $t = \frac{3\pi}{4}$ and $t = \frac{7\pi}{4} = \frac{3\pi}{4} + \pi$ are the same, with both giving $\sin(2t) = -1$, so $\frac{\sin(2t)}{2} = -\frac{1}{2}$ is a minimum at both points. Also, finally note that $\frac{5\pi}{4} \in [0,2\pi]$ and $\frac{7\pi}{4} \in [0,2\pi]$.
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