How do you find the critical numbers of $y = sin^2 x$ ?

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Answer 1 · 2016-03-03T23:13:23

We know that $cos2x=cos^2x-sin^2x=>cos2x=1-2*sin^2x=> sin^2x=1/2[1-cos2x]$

Hence $y$ becomes $y=1/2*[1-cos2x]$

Now the critical values $c$ of $f$ are those if and only if

i. $f'(c)=0$
ii. $f'(c)$ is undefined

So the critical values are those for which $dy/(dc)=0$ hence

$dy/(dc)=sin2c=0=>2c=n*pi=>c=n*pi/2$

where n is an integer.

The critical points are at $x=(n*pi)/2$

How do you find the critical numbers of y = sin^2 xy = sin^2 x?