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StokesParadox
 2 years ago
@zepdrix
StokesParadox
 2 years ago
@zepdrix

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zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm darn, I have plenty of resources for Calculus... I can't think of any for trig though haha.

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0I'm looking at yesterdays problems...one seconds.... we should probably start with inverses of trig functions..one second

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/SolveTrigEqn.aspx

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0do these look difficult enough (for me)

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Hmm they look pretty simple. I wouldn't say they're way easier than what we were working on yesterday. They might be worth spending some time on c:

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0yeah they look too easy....I'll find some harder ones \[sin(5x)=\frac{\sqrt 3}{2}\] dw:1362793858479:dw am I right so far?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1I mean, yes you're right so far!

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Keep in mind we want to rotate in the positive direction. So make sure your angle corresponds to that.

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0\[sin^{1}(\sqrt{3}/2)=5x\] uhm I'm stuck \[\frac {5\pi} 3\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large 5\pi/3=5x\] So you found that your angle, \(\large 5x\) was equal to the special angle \(\large 5\pi/3\). Looks good so far.

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0oh so that is right!

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0\[x=\frac {\pi}{3}\]

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0wait, that's it? what did we just discover?

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0wait wait wait....

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1There is some angle \(\large 5x\) that corresponded to the given equation. And we were ultimately trying to find 1/5 of that angle. No waiting! >:O lol

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0LOL \[sin(5x)=\frac{\sqrt 3}{2}\] why are we trying to find 1/5 of that angle do you mean: "there is some angle x"

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Our angle is the thing inside of the trig function. Think of it maybe as like, \(\large 5x=\theta\). We want to solve for part of that angle. But for the purposes of relating this to special angles, it's good to think of the entire contents as the angle. :)

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0I am.....so why are we trying to find 1/5 of that angle? Oh you're just saying we solve for x which is 1/5th of that angle....I think I get it

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1si c: I'm just being a little technical I guess.

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0It makes sense :)

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0should we do a more difficult problem? what are these problems called \[sec\left(cos^{1}\frac 12\right)\] I feel sooooo d u m b LOL

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Ok so we're dealing with an inverse even function on the inside. So our angle will be in the 1st or second quadrant, yes? Hmmm +1/2. Is that a length to the right, or left? :D

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0I will feel smarter by the end of the weekend....hopefully.... When I Know How To Do These Darn Problems o.O

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0the length to the right

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0I found some nice aerobics music!!!!

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0I'm planning to work out in 15 mins

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0I just need to find a way to download it

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0....so these problems are called inverse functions within trig functions? i"m trying to find some problems on google.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry I'm not quite sure what they're called :( Maybe something Composition of Trig functions...

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1http://www.ck12.org/book/CK12TrigonometryConcepts/r1/section/4.6/ Near the bottom there is a \(\large \text{Practice}\) section. Some good problems there.

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0which one should a take a stab at?

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0\[tan(cos^{1}1)\] \[cos\alpha =1\] at zero and \(2\pi\)

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1\[\large \tan(0)\] K looks good so far.

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0hmmmmm..... 0? final answer

StokesParadox
 2 years ago
Best ResponseYou've already chosen the best response.0yayyayayayayay!!!!! ok I'm gonna work out now. THanks!!!
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