## anonymous one year ago If sin theta=3/5 and theta is in quad 2 the exact form of csc (pi/2-theta) is...? Just want to make sure if this works??

1. anonymous

sinθ = 3/5 csc(π/2 - θ) = secθ = 1/cosθ = -4/5

2. triciaal

|dw:1436736243879:dw|

3. triciaal

given sin = 3/5 and in quad 2

4. anonymous

@triciaal So what does that mean?

5. triciaal

csc = 1/sin = hypotenuse / opposite 5/-4

6. anonymous

7. triciaal

|dw:1436737161908:dw|

8. anonymous

What do I do with that?

9. triciaal

your question "the exact form of csc (pi/2-theta) is?"

10. triciaal

shouldn't it be between -1 and 1 only?

11. freckles

"sinθ = 3/5 csc(π/2 - θ) = secθ = 1/cosθ = -4/5" in your work here the only mistake you made was not flipping -4/5 that cos(theta)=-4/5 so 1/cos(theta) is the reciprocal of that

12. triciaal

csc = hypotenuse/ opposite

13. ybarrap

|dw:1436737547332:dw| However, since we are in the 2nd quadrant $$-\csc\left (\frac{\pi}{2}-\theta\right )=\frac{{\bf\color{red}{-}}5}{~~4}$$ Note the negative corresponding to the value of $$x$$ in the 2nd quadrant.

14. triciaal

@zepdrix why/when is the value < -1 in general?

15. zepdrix

Trish :) Maybe you're thinking about sine and cosine, which have restricted output values between 1 and -1

16. zepdrix

That's the only way they will satisfy a triangle hypotenuse longer than other sides$\large\rm \sin \theta=\frac{opposite}{hypotenuse}=\frac{shorter}{longer}\lt1$But for Cosecant, I guess we have the reverse going on, ya? We always need to be 1 or larger.$\Large\rm \csc \theta=\frac{hypotenuse}{opposite}=\frac{longer}{shorter}\gt1$

17. ybarrap

Does this make sense? |dw:1436738761632:dw|

18. triciaal

@zepdrix thanks just never thought about it @ybarrap thanks for the CAST diagram but I was checking the limit ; not if positive or negative

19. triciaal

@ybarrap no tan on your chart

20. ybarrap

|dw:1436739672137:dw|

21. anonymous

Sorry guys I'm back! So i think I got the answer this time around! @triciaal @zepdrix

22. anonymous

-1/3/5 I think @zepdrix @triciaal @freckles

23. freckles

what is that the answer to?

24. freckles

are you looking at a new question?

25. anonymous

No for the one I put on

26. freckles

oh I guess you didn't understand what anyone posted :p

27. anonymous

Yeah nothing at all XD

28. anonymous

This is the only question I can't get!! :////

29. freckles

ok let's start from the beginning this shouldn't be long at all

30. freckles

"sinθ = 3/5 csc(π/2 - θ) = secθ = 1/cosθ = -4/5" do you remember this is what you first posted here?

31. freckles

this was your first try here anyways

32. anonymous

Yeah thats what I did, but I figured it was wrong

33. freckles

it is a little wrong

34. freckles

the first part is totally awesome

35. freckles

the second part the part where you say 1/cos(theta)=-4/5 is a little off

36. freckles

yes cos(theta)=-4/5 but what is 1/cos(theta)=?

37. freckles

if I tell you a=2/5 then what is 1/a=?

38. freckles

39. freckles

a and 1/a are reciprocals of one another that means they are the "flippings of each other" like 2 and 1/2 are reciprocals and 2/5 and 5/2 are reciprocals and -2/5 and -5/2 are reciprocals

40. freckles

so if I say a=2/5 then 1/a=5/2 or if I say 1/a=5/2 then a=2/5

41. freckles

so if cos(theta)=-4/5 then 1/cos(theta)=?

42. freckles

if you want to replace cos(theta) with a you can if a=-4/5 then 1/a=?

43. anonymous

@freckles Alright I think I need a little bit to take this all in cuz you just shot me a whole bunch of stuff at me XD I think I get what your saying let me see what I can do

44. freckles

45. anonymous

Wait so would it be -5/4 then??? *fingers crossed right now that its right** haha

46. freckles

yes yes !!!

47. anonymous

YYAYYY!!

48. freckles

-5/4 is indeed the reciprocal of -4/5

49. anonymous

So where do you go on from there? Is there more to be done??

50. freckles

nope

51. anonymous

Ohmygod that simple for a question like this? My mind is just blown right now .... XD

52. freckles

$"\sinθ = 3/5 \\ \csc(π/2 - θ) = \secθ = 1/\cosθ = -4/5" \\ \text{ so the correcting your work would look like this: } \\ \csc(\frac{\pi}{2}-\theta)=\sec(\theta)=\frac{1}{\cos(\theta)}=\frac{1}{\frac{-4}{5}}=\frac{-5}{4}$

53. freckles

by the way you know that first thing you used it called a cofunction identity most people forget that one so you are doing awesome!

54. anonymous

@freckles Thanks again seriously!! Haha really helped! Literally am beating myself up not knowing lol

55. freckles

i think we were all working on our algebra in trig and even in calculus

56. freckles

algebra/arithemetic

57. freckles

and/or

58. freckles

guess what you are going to have to know so much algebra for calculus you are going to think you never learned algebra :p well that has been my experience with some anyways I'm kind a kidding. I know you are going do okay especially since you can always get help if you need it.

59. anonymous

Haha true!! thanks again! I know with Calc I am going to need lots of help especially!!

60. triciaal

math is fun just takes lots of practice