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

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

StokesParadoxBest ResponseYou've already chosen the best response.0
http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/SolveTrigEqn.aspx
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
do these look difficult enough (for me)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Hmm 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:
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.1
I mean, yes you're right so far!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Keep in mind we want to rotate in the positive direction. So make sure your angle corresponds to that.
 one year ago

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

zepdrixBest 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.
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
oh so that is right!
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
\[x=\frac {\pi}{3}\]
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
wait, that's it? what did we just discover?
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
wait wait wait....
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
There 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
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.1
Our 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. :)
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
I 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
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
si c: I'm just being a little technical I guess.
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
It makes sense :)
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
should 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
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Ok 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
 one year ago

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

StokesParadoxBest ResponseYou've already chosen the best response.0
the length to the right
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
I found some nice aerobics music!!!!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
lol that was random XD
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
I'm planning to work out in 15 mins
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
I just need to find a way to download it
 one year ago

StokesParadoxBest 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.
 one year ago

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

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

StokesParadoxBest ResponseYou've already chosen the best response.0
which one should a take a stab at?
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.1
\[\large \tan(0)\] K looks good so far.
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
hmmmmm..... 0? final answer
 one year ago

StokesParadoxBest ResponseYou've already chosen the best response.0
yayyayayayayay!!!!! ok I'm gonna work out now. THanks!!!
 one year ago
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