Question

@zepdrix

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

do these look difficult enough (for me)

Answer 5

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:

Answer 6

Take a stab at #3

Answer 7

yeah they look too easy....I'll find some harder ones
\[sin(5x)=-\frac{\sqrt 3}{2}\]

am I right so far?

Answer 8

Solve for x! :)

Answer 9

I mean, yes you're right so far!

Answer 10

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

Answer 11

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

Answer 12

\[\large 5\pi/3=5x\]

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

Answer 13

oh so that is right!

Answer 14

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

Answer 15

Yay good job

Answer 16

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

Answer 17

wait wait wait....

Answer 18

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

Answer 19

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"

Answer 20

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. :)

Answer 21

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

Answer 22

si?

Answer 23

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

Answer 24

It makes sense :)

Answer 25

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

Answer 26

lol poor gal c:

Answer 27

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

Answer 28

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

Answer 29

right

Answer 30

the length to the right

Answer 31

I found some nice aerobics music!!!!

Answer 32

lol that was random XD

Answer 33

Thanks :)

Answer 34

I'm planning to work out in 15 mins

Answer 35

I just need to find a way to download it

Answer 36

Oh i see c:

Answer 37

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

Answer 38

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

Answer 39

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.

Answer 40

which one should a take a stab at?

Answer 41

5

Answer 42

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

Answer 43

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

Answer 44

hmmmmm.....

0? final answer

Answer 45

Yes good!

Answer 46

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

Answer 47

Cya c: