- jamiebookeater

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- zepdrix

Hmm darn, I have plenty of resources for Calculus... I can't think of any for trig though haha.

- anonymous

I'm looking at yesterdays problems...one seconds....
we should probably start with inverses of trig functions..one second

- anonymous

http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/SolveTrigEqn.aspx

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## More answers

- anonymous

do these look difficult enough (for me)

- zepdrix

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:

- zepdrix

Take a stab at #3

- anonymous

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?

- zepdrix

Solve for x! :)

- zepdrix

I mean, yes you're right so far!

- zepdrix

Keep in mind we want to rotate in the positive direction. So make sure your angle corresponds to that.

- anonymous

\[sin^{-1}(-\sqrt{3}/2)=5x\]
uhm I'm stuck
\[\frac {5\pi} 3\]

- zepdrix

\[\large 5\pi/3=5x\]
So you found that your angle, \(\large 5x\) was equal to the special angle \(\large 5\pi/3\).
Looks good so far.

- anonymous

oh so that is right!

- anonymous

\[x=\frac {\pi}{3}\]

- zepdrix

Yay good job

- anonymous

wait, that's it? what did we just discover?

- anonymous

wait wait wait....

- zepdrix

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

- anonymous

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"

- zepdrix

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. :)

- anonymous

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

- anonymous

si?

- zepdrix

si c: I'm just being a little technical I guess.

- anonymous

It makes sense :)

- anonymous

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

- zepdrix

lol poor gal c:

- zepdrix

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

- anonymous

I will feel smarter by the end of the weekend....hopefully....
When I Know How To Do These Darn Problems o.O

- anonymous

right

- anonymous

the length to the right

- anonymous

I found some nice aerobics music!!!!

- zepdrix

lol that was random XD

- anonymous

Thanks :)

- anonymous

I'm planning to work out in 15 mins

- anonymous

I just need to find a way to download it

- zepdrix

Oh i see c:

- anonymous

....so these problems are called inverse functions within trig functions? i"m trying to find some problems on google.

- zepdrix

Sorry I'm not quite sure what they're called :(
Maybe something Composition of Trig functions...

- zepdrix

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.

- anonymous

##### 1 Attachment

- anonymous

which one should a take a stab at?

- zepdrix

5

- anonymous

\[tan(cos^{-1}1)\]
\[cos\alpha =1\]
at zero and \(2\pi\)

- zepdrix

\[\large \tan(0)\]
K looks good so far.

- anonymous

hmmmmm.....
0? final answer

- zepdrix

Yes good!

- anonymous

yayyayayayayay!!!!! ok I'm gonna work out now. THanks!!!

- zepdrix

Cya c:

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