1. ## Even or odd

Hey guys.

I've got this function.

g (x) = 1 - cos (pi - 2x)

I must prove this is an even function.

All I got was g (-x) = 1 - cos (pi + 2x), but I can't pass this point. Where do I go from here? Any help would be appreciated. Thank you.

2. Originally Posted by Alvy
Hey guys.

I've got this function.

g (x) = 1 - cos (pi - 2x)

I must prove this is an even function.

All I got was g (-x) = 1 - cos (pi + 2x), but I can't pass this point. Where do I go from here? Any help would be appreciated. Thank you.
Hi Alvy,

use the fact that

$cos({\pi}-2x)=-cos(-2x)$

$cos(-2x)=cos(2x)$

$cos({\pi}+2x)=-cos(2x)$

3. Originally Posted by Alvy
Hey guys.

I've got this function.

g (x) = 1 - cos (pi - 2x)

I must prove this is an even function.

All I got was g (-x) = 1 - cos (pi + 2x), but I can't pass this point. Where do I go from here? Any help would be appreciated. Thank you.
hi

this is an even function . If you graph it , you will see that the function is symmetrical about the y-axis .

4. Originally Posted by Archie Meade
Hi Alvy,

use the fact that

$cos({\pi}-2x)=-cos(-2x)$
sorry , understood .

5. Thank you very much, Archie Meade and mathaddict. I got it. One last question, if I may.

I must find g(x), knowing that sen(2x) = 1/3, and x E [pi/4, pi/2].

What's the path here? I'm confused. Thank you.

6. Originally Posted by Alvy
Thank you very much, Archie Meade and mathaddict. I got it. One last question, if I may.

I must find g(x), knowing that sen(2x) = 1/3, and x E [pi/4, pi/2].

What's the path here? I'm confused. Thank you.
$sin(2x)=\frac{1}{3}$

$2x=sin^{-1}\left(\frac{1}{3}\right)=19.47^o$

This is too small given the domain.

However $sin(2x)=sin({\pi}-2x)$

Hence $180^o-2x=19.47^o$

This gives a valid x.