Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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:
Take a stab at #3
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?
Solve for x! :)
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!
\[x=\frac {\pi}{3}\]
Yay good job
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?
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
lol poor gal c:
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
right
the length to the right
I found some nice aerobics music!!!!
lol that was random XD
Thanks :)
I'm planning to work out in 15 mins
I just need to find a way to download it
Oh i see c:
....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?
5
\[tan(cos^{-1}1)\] \[cos\alpha =1\] at zero and \(2\pi\)
\[\large \tan(0)\] K looks good so far.
hmmmmm..... 0? final answer
Yes good!
yayyayayayayay!!!!! ok I'm gonna work out now. THanks!!!
Cya c:

Not the answer you are looking for?

Search for more explanations.

Ask your own question