evaluating the following.

**jvignacio** · September 11th 2008, 10:43 PM

evaluating
$ sin(\frac{-9\pi}{4}) $

just wondering how they got

$ \frac{-1}{\sqrt{2}} $

i got

$ \frac{-\sqrt{2}}{2} $
but was wrong

**11rdc11** · September 11th 2008, 10:59 PM

Try this

$\bigg(\frac{-\sqrt{2}}{2}\bigg)\bigg(\frac{\sqrt{2}}{\sqrt{2}}\ bigg)$

$\frac{-2}{2\sqrt{2}}$

$\frac{-1}{\sqrt{2}}$

**justinwager** · September 11th 2008, 11:00 PM

I got 2 root 2 as well..isn't this in the 1st quadarant...hence Sin = a positive...

**11rdc11** · September 11th 2008, 11:03 PM

Its in the 4th quadarant

**jvignacio** · September 11th 2008, 11:04 PM

Originally Posted by justinwager

I got 2 root 2 as well..isn't this in the 1st quadarant...hence Sin = a positive...

yeah thats what i thought.. its wierd. dont know how they got the -1 and root 2 on the denom

**jvignacio** · September 11th 2008, 11:05 PM

Originally Posted by 11rdc11

Its in the 4th quadarant

yeah coz u go backyards , but it still shud be what i wrote..

**jvignacio** · September 11th 2008, 11:09 PM

Originally Posted by 11rdc11

Try this

$\bigg(\frac{-\sqrt{2}}{2}\bigg)\bigg(\frac{\sqrt{2}}{\sqrt{2}}\ bigg)$

$\frac{-2}{2\sqrt{2}}$

$\frac{-1}{\sqrt{2}}$

why did u multiply it by 1 ? sqrt 2 / sqrt 2

**11rdc11** · September 11th 2008, 11:10 PM

To make it look like the answer provided. It doesn't change the answer since what I did to the numerator I also did to the denominator so it valid

$\frac{-\sqrt{2}}{2} = \frac{-1}{\sqrt{2}} $

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