# Thread: Help with Trig Identities exercise.

1. ## 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.

2. ## Re: Help with Trig Identities exercise.

Note:

$\displaystyle \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{\sqrt{9}} = \frac{2\sqrt{2}}{3}$.

3. ## 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.

4. ## Re: Help with Trig Identities exercise.

Do you know that $\displaystyle \sqrt{4}= 2$? And 8= (4)(2).

5. ## 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.

6. ## Re: Help with Trig Identities exercise.

note that $\displaystyle \sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$

$\displaystyle \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2}$

7. ## Re: Help with Trig Identities exercise.

Ah, alright, I got it. Thank you everyone for your help.