# Max,min of sinx

• Apr 10th 2010, 12:23 PM
charikaar
Max,min of sinx
Hi,

I know sinx=1 for x=$\displaystyle \pi/2\pm\pi*2n$

and sinx=-1 for x=$\displaystyle \pi*1.5\pm\pi*2n$

How the above x values change if we have $\displaystyle \sqrt sinx$

thanks
• Apr 10th 2010, 12:33 PM
$\displaystyle \sqrt{|\sin x|}$ will have minima at the zeros of sine and maxima at the extrema of sine, because the square root is monotonic. I'm not sure what to do without those absolute value bars though....
• Apr 10th 2010, 02:02 PM
mathemagister
Quote:

Originally Posted by charikaar
Hi,

I know sinx=1 for x=$\displaystyle \pi/2\pm\pi*2n$

and sinx=-1 for x=$\displaystyle \pi*1.5\pm\pi*2n$

How the above x values change if we have $\displaystyle \sqrt sinx$

thanks

Quote:

$\displaystyle \sqrt{|\sin x|}$ will have minima at the zeros of sine and maxima at the extrema of sine, because the square root is monotonic. I'm not sure what to do without those absolute value bars though....

Well, you will notice that, as maddas said, you will have "minima at the zeros of sine and maxima at the extrema of sine." The absolute value bars will not be a problem because you don't really have to solve for anything inside them. Think about it logically... the answers will not be the same as for sinx, but twice as frequent (so you must divide the 2n(pi) by 2) because of the absolute value. So, the maxima will be

$\displaystyle \max(\sqrt{|\sin{x}|}) = 1 \quad \text{at} \quad x = \frac{\pi}{2} \pm n \pi$

Maybe you really did mean $\displaystyle \sqrt{\sin{x}}$, without the absolute value. Then, the maxima will be the same as before, since square rooting a graph will just make it a bit more "squished down," while keeping the overall structure. So, without the absolute value bars, the maxima will remain:

$\displaystyle \max(\sqrt{\sin{x}})=1 \quad \text{at} \quad x=\frac{\pi}{2} \pm 2n\pi$

Hope that helps :)

Mathemagister