evaluating $\displaystyle sin(\frac{-9\pi}{4}) $ just wondering how they got $\displaystyle \frac{-1}{\sqrt{2}} $ i got $\displaystyle \frac{-\sqrt{2}}{2} $ but was wrong
Follow Math Help Forum on Facebook and Google+
Try this $\displaystyle \bigg(\frac{-\sqrt{2}}{2}\bigg)\bigg(\frac{\sqrt{2}}{\sqrt{2}}\ bigg)$ $\displaystyle \frac{-2}{2\sqrt{2}}$ $\displaystyle \frac{-1}{\sqrt{2}}$
I got 2 root 2 as well..isn't this in the 1st quadarant...hence Sin = a positive...
Its in the 4th quadarant
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
Originally Posted by 11rdc11 Its in the 4th quadarant yeah coz u go backyards , but it still shud be what i wrote..
Originally Posted by 11rdc11 Try this $\displaystyle \bigg(\frac{-\sqrt{2}}{2}\bigg)\bigg(\frac{\sqrt{2}}{\sqrt{2}}\ bigg)$ $\displaystyle \frac{-2}{2\sqrt{2}}$ $\displaystyle \frac{-1}{\sqrt{2}}$ why did u multiply it by 1 ? sqrt 2 / sqrt 2
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 $\displaystyle \frac{-\sqrt{2}}{2} = \frac{-1}{\sqrt{2}} $
View Tag Cloud