Help with Trig Identities exercise.

**fogownz** · November 15th 2012, 08:43 AM

Ok, I'm supposed to find sin $\theta$ when cos $\theta$ = $\frac{1}{3}$ . The answer is $\frac{2\sqrt{2}}{3}$ . I got $\sqrt{\frac{8}{9}$ . Does this somehow simplify to the correct answer? it seems like it might, but my algebra skills are terrible.

**Aryth** · November 15th 2012, 08:58 AM

Note:

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

**fogownz** · November 15th 2012, 09:53 AM

Awesome, thank you.

Edit: Actually, I don't quite understand how it is determined that $\sqrt{8}=2\sqrt{2}$ I know that it does, but how could I have gotten there.

**HallsofIvy** · November 15th 2012, 11:57 AM

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

**fogownz** · November 15th 2012, 01:27 PM

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 $\sqrt{4}=2$ , yet $4\sqrt{2}$ does not equal $\sqrt{16}$ . There is something that I'm missing from this.

**skeeter** · November 15th 2012, 01:39 PM

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

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

**fogownz** · November 15th 2012, 01:47 PM

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

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