Help with Trig Identities exercise.

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November 15th 2012, 08:43 AM
fogownz

Help with Trig Identities exercise. (unsolved)

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.
November 15th 2012, 08:58 AM
Aryth

Re: Help with Trig Identities exercise.

Note:

$\sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{\sqrt{9}} = \frac{2\sqrt{2}}{3}$ .
November 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 $\sqrt{8}=2\sqrt{2}$ I know that it does, but how could I have gotten there.
November 15th 2012, 11:57 AM
HallsofIvy

Re: Help with Trig Identities exercise.

Do you know that $\sqrt{4}= 2$ ? And 8= (4)(2).
November 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 $\sqrt{4}=2$ , yet $4\sqrt{2}$ does not equal $\sqrt{16}$ . There is something that I'm missing from this.
November 15th 2012, 01:39 PM
skeeter

Re: Help with Trig Identities exercise.

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

$\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2}$
November 15th 2012, 01:47 PM
fogownz

Re: Help with Trig Identities exercise.

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

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