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## StokesParadox 2 years ago @zepdrix

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1. zepdrix

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

2. StokesParadox

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

3. StokesParadox
4. StokesParadox

do these look difficult enough (for me)

5. 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:

6. zepdrix

Take a stab at #3

7. StokesParadox

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?

8. zepdrix

Solve for x! :)

9. zepdrix

I mean, yes you're right so far!

10. zepdrix

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

11. StokesParadox

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

12. 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.

13. StokesParadox

oh so that is right!

14. StokesParadox

$x=\frac {\pi}{3}$

15. zepdrix

Yay good job

16. StokesParadox

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

17. StokesParadox

wait wait wait....

18. 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

19. StokesParadox

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"

20. 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. :)

21. StokesParadox

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

22. StokesParadox

si?

23. zepdrix

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

24. StokesParadox

It makes sense :)

25. StokesParadox

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

26. zepdrix

lol poor gal c:

27. 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

28. StokesParadox

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

29. StokesParadox

right

30. StokesParadox

the length to the right

31. StokesParadox

I found some nice aerobics music!!!!

32. zepdrix

lol that was random XD

33. StokesParadox

Thanks :)

34. StokesParadox

I'm planning to work out in 15 mins

35. StokesParadox

I just need to find a way to download it

36. zepdrix

Oh i see c:

37. StokesParadox

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

38. zepdrix

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

39. 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.

40. StokesParadox

41. StokesParadox

which one should a take a stab at?

42. zepdrix

5

43. StokesParadox

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

44. zepdrix

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

45. StokesParadox

hmmmmm..... 0? final answer

46. zepdrix

Yes good!

47. StokesParadox

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

48. zepdrix

Cya c:

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