Help with Trig Identities exercise.

• Nov 15th 2012, 08:43 AM
fogownz
Help with Trig Identities exercise. (unsolved)
Ok, I'm supposed to find sin$\displaystyle \theta$ when cos$\displaystyle \theta$ = $\displaystyle \frac{1}{3}$. The answer is $\displaystyle \frac{2\sqrt{2}}{3}$. I got $\displaystyle \sqrt{\frac{8}{9}$. Does this somehow simplify to the correct answer? it seems like it might, but my algebra skills are terrible.
• Nov 15th 2012, 08:58 AM
Aryth
Re: Help with Trig Identities exercise.
Note:

$\displaystyle \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{\sqrt{9}} = \frac{2\sqrt{2}}{3}$.
• Nov 15th 2012, 09:53 AM
fogownz
Re: Help with Trig Identities exercise.
Awesome, thank you.

Edit: Actually, I don't quite understand how it is determined that $\displaystyle \sqrt{8}=2\sqrt{2}$ I know that it does, but how could I have gotten there.
• Nov 15th 2012, 11:57 AM
HallsofIvy
Re: Help with Trig Identities exercise.
Do you know that $\displaystyle \sqrt{4}= 2$? And 8= (4)(2).
• Nov 15th 2012, 01:27 PM
fogownz
Re: Help with Trig Identities exercise.
Yes, but that doesn't really click for me. When I saw that, I tried applying the steps I saw to a different problem. (4)(4)=16, and $\displaystyle \sqrt{4}=2$, yet $\displaystyle 4\sqrt{2}$ does not equal $\displaystyle \sqrt{16}$. There is something that I'm missing from this.
• Nov 15th 2012, 01:39 PM
skeeter
Re: Help with Trig Identities exercise.
note that $\displaystyle \sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$

$\displaystyle \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2}$
• Nov 15th 2012, 01:47 PM
fogownz
Re: Help with Trig Identities exercise.
Ah, alright, I got it. Thank you everyone for your help.