anonymous
  • anonymous
sin(arcsin(-11π/13) I'd really just like an explanation of this one.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I don't know why I'm forgetting how to do this, but it's been awhile and my final is coming up and I want a better understanding of it.
anonymous
  • anonymous
do you know the init Circle?
anonymous
  • anonymous
Yep

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anonymous
  • anonymous
when you at look the unit circle what quard would Sin be in?
anonymous
  • anonymous
I see that it goes -11π, so I believe it's going clockwise around the unit circle and I think it gets to Q3.
anonymous
  • anonymous
yes correct, so in the problem what does that tell you?
anonymous
  • anonymous
I don't know...hahaha, rephrase that maybe?
anonymous
  • anonymous
from looking at this problem and taking the information you already have what would you say the answer is? sin(arcsin(-11π/13)
anonymous
  • anonymous
do you understand arcsin and what that means??
anonymous
  • anonymous
crap, I wrote it down wrong. It was arcsin(sin(-11π/13))
anonymous
  • anonymous
I have a pretty good understanding of how to use it. I'm rusty, though.
anonymous
  • anonymous
okay, let us take this and break it down on what you know, you start I will fill in if needed.
anonymous
  • anonymous
I believe sin(ø) = arcsin (-11π/13) in this situation.
anonymous
  • anonymous
you are right, but how did you do that?
anonymous
  • anonymous
I just kind of have that memorized. I guess I never really grasped that...
anonymous
  • anonymous
okay, now I have a stupid what is the question asking for??? I am a little lost with the question here does it ask to solve or what?
anonymous
  • anonymous
Find the exact value of the given expression. Simplify your answer (rationalize the denom)
anonymous
  • anonymous
I also have that the answer is -2π/13
anonymous
  • anonymous
I don't really understand where the 2π comes from.
anonymous
  • anonymous
because recall the unit circle has a total of 360 degrees right?
anonymous
  • anonymous
Yep and 2π is 360º
anonymous
  • anonymous
2/pie is a point on the Unit the circle in which the angle is 360.. you beat to that one good job
anonymous
  • anonymous
So, my picture shows the line -11π/13 and the distance between that line and the 180º line is 2π/13.
anonymous
  • anonymous
Having a hard time understanding why it's 2π and not just π.
anonymous
  • anonymous
Probably just overthinking it.
anonymous
  • anonymous
what I am questioning here the negative here in this case?
anonymous
  • anonymous
because you start with a negative you will most likely end that way..
anonymous
  • anonymous
2pie is at the 360 degrees which has points 0,0 so how can that be a negative result help here?
anonymous
  • anonymous
One sec, gonna post a photo of what is giving me a headache.
anonymous
  • anonymous
I think we are both getting the same headache on this. This problem is going down..
anonymous
  • anonymous
Hahaha, I'm tellin' you. I haven't had any problems with any other questions, but this one is just destroying my brain.
anonymous
  • anonymous
yeah, the same thing only that my brain is dead because I am stuying Vectors myself.
anonymous
  • anonymous
what I am missing here????
lgbasallote
  • lgbasallote
is the question still \[\sin (arcsin (-11 \pi/3)\]
anonymous
  • anonymous
arcsin(sin(-11π/3))
anonymous
  • anonymous
yes sir, please help me I am getting a major headache here
anonymous
  • anonymous
Messed up the on the original post
lgbasallote
  • lgbasallote
here's a hint.. arcsin( sin x) = x sin (arcsin x) = x cos( arc cos x) = x arc cos (cos x) = x tan (arctan x) = x arctan (tan x) = x does that help?
anonymous
  • anonymous
anonymous
  • anonymous
There is the picture that I was trying to describe.
anonymous
  • anonymous
I wish that helped, but I still can't figure out why it is -2π/13
lgbasallote
  • lgbasallote
is that included in the question? or is that your solytion?
anonymous
  • anonymous
Solution. It's an answer key and I'm trying to decipher it.
anonymous
  • anonymous
I know it is -2π/13 because of the range, but I don't know where the 2π came from.
lgbasallote
  • lgbasallote
use the hint i gave you...here's a demo \[\sin (\sin^{-1} ( \frac{\pi}{2} ) ) = \frac{\pi}{2}\] \[\sin^{-1} (\sin ( \frac{2 \pi}{5} ) ) = \frac{2 \pi}{5}\] getting it now?
lgbasallote
  • lgbasallote
\[\sin^{-1} (\sin ( \frac{-11\pi}{3} ) ) = -\frac{11\pi}{3}\] then simplify that you get \[-\frac{2\pi}{3}\] make sense?
anonymous
  • anonymous
lgbasallote what did I miss here? I mean I thought for sure I was going the right way what did I do in my steps that went wrong here? Help me to understand..
anonymous
  • anonymous
How'd you simplify to that?
lgbasallote
  • lgbasallote
i think i wrote the wrong thingies lol...
lgbasallote
  • lgbasallote
but that's the concept
anonymous
  • anonymous
Yeah, I knew what you were saying with the others. It's this deal here. I don't understand how it is simplifying from 11π to 2π.
anonymous
  • anonymous
It's like it just magically disappears.
lgbasallote
  • lgbasallote
reference angles
anonymous
  • anonymous
Heh, I feel dumb for not knowing still.
anonymous
  • anonymous
Mrcarey I am so sorry I took the long way here.
anonymous
  • anonymous
Haha, no problem at all. I've been at it for so long, it didn't even feel that long.
anonymous
  • anonymous
ha ha yeah better next time right? HA HA..
anonymous
  • anonymous
Exactly! Thanks for trying at least. Good luck with the vector calc. I'll be THERE soon enough.
anonymous
  • anonymous
I am Physics Engineering major myelf
anonymous
  • anonymous
Mech E. for me. Math math math math and more math!
anonymous
  • anonymous
you got this now
anonymous
  • anonymous
Yeah, I feel better about it. Got one more day of review in class, so I'll probably bring it up with him anyway. Thanks again!
anonymous
  • anonymous
do you another problem you need help on? Just asking here
anonymous
  • anonymous
Nope, this one seemed to be the only one I kept coming back to. This stupid thing made everything else seem easy!
anonymous
  • anonymous
yeah, I only hope for me it is just one problem that will allow everything to be easy after that. Fat chance ha ha
anonymous
  • anonymous
Hahahaha, no way. I've peeked at that stuff and needed to take a nap after.
anonymous
  • anonymous
did that one already.. I took a 5hrs nap
anonymous
  • anonymous
I am doing Calc1 andPhysics 1 with Calc
anonymous
  • anonymous
I'll be doing that in the Fall. Well, no Physics yet, couldn't get in the class. General Chem 1 and Calc 1. Gonna be exciting.
anonymous
  • anonymous
what school?
anonymous
  • anonymous
Doing community for now, so I can transfer to University of Washington Seattle.
anonymous
  • anonymous
You?
anonymous
  • anonymous
cool I live Iin Portland Ore myself..
anonymous
  • anonymous
I will start Sept 24th at Jr College than transfer to Tarleton ST university in TX in Spring 2013
anonymous
  • anonymous
Awesome, TX should be pretty fun. Definitely a change of scenery.
anonymous
  • anonymous
I am studying math and Physics by watching videos and taking notes on my own
anonymous
  • anonymous
Khan Academy?
anonymous
  • anonymous
yeah StephenVille is very small town though
anonymous
  • anonymous
who is Khan A?
anonymous
  • anonymous
www.khanacademy.com MIT grad who made his own site that teaches every kind of math and most sciences.
anonymous
  • anonymous
Free site for learning. I've used it for awhile now. Especially when I get bad professors.
anonymous
  • anonymous
funny you just posted MIT, Because I am using thier open courses on line for the video in math and Physics
anonymous
  • anonymous
Haha, that works, too! You'd probably like his teaching style. He is very helpful and has tons of videos. I always recommend the site.
anonymous
  • anonymous
thanks will check it out, hey you have a great night and if you need me to help me message me
anonymous
  • anonymous
you
anonymous
  • anonymous
Sounds like a plan! Have a good night!
anonymous
  • anonymous
you too.

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