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StokesParadox
@zepdrix
Hmm darn, I have plenty of resources for Calculus... I can't think of any for trig though haha.
I'm looking at yesterdays problems...one seconds.... we should probably start with inverses of trig functions..one second
http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/SolveTrigEqn.aspx
do these look difficult enough (for me)
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:
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?
I mean, yes you're right so far!
Keep in mind we want to rotate in the positive direction. So make sure your angle corresponds to that.
\[sin^{-1}(-\sqrt{3}/2)=5x\] uhm I'm stuck \[\frac {5\pi} 3\]
\[\large 5\pi/3=5x\] So you found that your angle, \(\large 5x\) was equal to the special angle \(\large 5\pi/3\). Looks good so far.
oh so that is right!
\[x=\frac {\pi}{3}\]
wait, that's it? what did we just discover?
wait wait wait....
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
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"
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. :)
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
si c: I'm just being a little technical I guess.
It makes sense :)
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
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
I will feel smarter by the end of the weekend....hopefully.... When I Know How To Do These Darn Problems o.O
the length to the right
I found some nice aerobics music!!!!
I'm planning to work out in 15 mins
I just need to find a way to download it
....so these problems are called inverse functions within trig functions? i"m trying to find some problems on google.
Sorry I'm not quite sure what they're called :( Maybe something Composition of Trig functions...
http://www.ck12.org/book/CK-12-Trigonometry-Concepts/r1/section/4.6/ Near the bottom there is a \(\large \text{Practice}\) section. Some good problems there.
which one should a take a stab at?
\[tan(cos^{-1}1)\] \[cos\alpha =1\] at zero and \(2\pi\)
\[\large \tan(0)\] K looks good so far.
hmmmmm..... 0? final answer
yayyayayayayay!!!!! ok I'm gonna work out now. THanks!!!