## kittiwitti1 one year ago Problems which I cannot for the life of me figure out why I got wrong/partially incorrect: A: Find all six trigonometric functions of θ if the given point is on the terminal side of θ. (Assume that a is a positive number. If an answer is undefined, enter UNDEFINED.) (−5a, −12a) Result: http://prntscr.com/8diac1 B: Use your calculator to find sin θ and cos θ if the point (9.37, 7.05) is on the terminal side of θ. (Round your answers to four decimal places.) Result: http://prntscr.com/8dia9h C: Indicate the quadrants in which the terminal side of θ must lie under the following conditions. (Select all that apply.) cos θ and cot θ have the same sign Result: http://prntscr.com/8diaew

1. anonymous

|dw:1441605814325:dw| So we get $\sin(\theta)=\sin(180+x)=-\sin(x)$ $\cos(\theta)=\cos(180+x)=-\cos(x)$$\tan(\theta)=\tan(180+x)=\tan(x)$ from the triangle we can get $\sin(x)=\frac{-12a}{\sqrt{(-12a)^2+(-5a)^2}}=\frac{-12a}{\sqrt{144a^2+25a^2}}=\frac{-12a}{13a}=-\frac{12}{13}$ Similarly find cos(x) and tan(x) and use them to find sin(theta), cos(theta) and tan(theta) reciprocal for cosec(theta) sec(theta) and cot(theta)

2. kittiwitti1

Which problem are you doing? (Sorry I'm multitasking) e_e

3. anonymous

Do you have a figure?Talking about problem 1

4. kittiwitti1

I'm only wondering where I went wrong, I don't need the whole process explained e_e;;;

5. kittiwitti1

Besides they said the sin was correct... My issue is how the tan/cot are incorrect.

6. anonymous

I see your problem now, what you did was, you found the ratios $\sin(x),\cos(x),\tan(x)$ BUT you're suppose to find $\sin(\theta),\cos(\theta),\tan(\theta)$ and NOT $\sin(x),\cos(x),\tan(x)$ what I mean by x, refer to the figure

7. kittiwitti1

I don't see what you mean by x... yes I looked at the figure. x_x;

8. anonymous

You did something like |dw:1441606633005:dw|

9. kittiwitti1

Yes but aren't the trig values the same?

10. anonymous

Nope!! It makes a slight difference

11. kittiwitti1

It does? *trig function values

12. anonymous

Refer to these formulae I've given below $\sin(\theta)=\sin(180+x)=-\sin(x)$$\cos(\theta)=\cos(180+x)=-\cos(x)$$\tan(\theta)=\tan(180+x)=\tan(x)$

13. anonymous

As you can see sign changes for sine and cosine but remains same for tangent

14. anonymous

Let's find your tan theta, since it's seem to be giving you problem $\tan(\theta)=\tan(x)=\frac{-12a}{-5a}=\frac{12}{5}$

15. kittiwitti1

Er. Yes...

16. anonymous

17. anonymous

that's why you're getting wrong answer

18. anonymous

and if your tangent is wrong then obviously cotangent will also be wrong

19. anonymous

notice you answers you've done $\frac{5}{12}$ when it should be $\frac{12}{5}$

20. anonymous

@kittiwitti1 you there ??

21. kittiwitti1

Sorry, working on other problems... What'd I miss?

22. anonymous

23. kittiwitti1

-facedesks- Dx I feel dumb. Okay, thanks; what about the other two?

24. anonymous

problem 3) Quadrant 2, both cos and cot have negative sign so that's ok Quadrant 3, cos is negative while cot is POSITIVE, so that's wrong how about Quadrant 1?both cos and cot, infact all functions are positive!

25. kittiwitti1
26. anonymous

|dw:1441607496705:dw|

27. freckles

$\sin(\theta)=\frac{y}{r} \text{ and } \cos(\theta)=\frac{x}{r} \\ \text{ and you have } r=\sqrt{x^2+y^2} \\ \text{ you can use this \to help you evaluate } \sin(\theta) \text{ and } \cos(\theta)$

28. kittiwitti1

Oh... I seem to have attracted quite the audience... sorry for previous "mass-tags" ^^;

29. anonymous

|dw:1441607639710:dw|

30. kittiwitti1

Wait what are you going on about lol

31. anonymous

32. anonymous

It's an easy way to remember

33. anonymous

like soh cah toa

34. kittiwitti1

Yes but what does that have to do with problem B o_o (I mean thanks for the info but I don't see the connection)

35. anonymous

In problem 3 quadrant is wrong look at it again

36. anonymous

37. anonymous

is wrong

38. anonymous

I told you about that just in case you have any trouble remembering what is positive in which quadrant

39. kittiwitti1

Oh okay that was the answer for problem 3, but what about #2 o_o?

40. anonymous

41. kittiwitti1

I was looking at my notes... did the teacher teach us the wrong things D:

42. kittiwitti1

Is it Q1-Q3?

43. freckles

cosine and cotangent have the same sign that means they could both be positive or they could both be negative if cosine and cotangent are both negative then what is sine going to be? if cosine and cotangent are both positive then what is sine going to be?

44. kittiwitti1

So Q1-Q3?

45. kittiwitti1

Okay

46. anonymous

Q3 is not the correct option as I said earlier

47. kittiwitti1

Wait what?

48. anonymous

in Q3 cos is negative while cot is positive not the same sign at all

49. kittiwitti1

My notes say Q3 has both negative values for cos and cot.

50. freckles

cosine and cotangent have the same sign that means they could both be positive or they could both be negative if cosine and cotangent are both negative then what is sine going to be? then sine is positive if cosine and cotangent are both positive then what is sine going to be? then sine is positive yes he is right!

51. freckles

@Nishant_Garg is right

52. freckles

I should have answered my own questions

53. kittiwitti1

I just realized I wrote a wrong trig function in my notes...

54. kittiwitti1

the* wrong Nevermind. What about problem B?

55. anonymous

you've written wrong dear. Take it from me, never refer to notes for absolute answers because we students tend to write notes during a class while trying to fight off sleep, not very accurate for your notes huh? :P

56. kittiwitti1

Lol but I already noticed before you'd said that. Actually the teacher had miswritten. I should have read the book... forgot the teacher always makes mistakes during lectures (she took 8 minutes till she realized she was going over the wrong problem once)

57. kittiwitti1

Even though we tried to tell her five times... What about problem B? http://prntscr.com/8dia9h

58. anonymous

Ouch, for problem 2 you can do the following remember the unit circle? we have the following equations $x=r\cos(\theta), y=r\sin(\theta)$ Dividing these we get $\frac{y}{x}=\frac{r\sin(\theta)}{r\cos(\theta)}=\tan(\theta)$ $\therefore \tan(\theta)=\frac{y}{x}=\frac{7.05}{9.37}$ Now that you've found 1 trignometric ratio, you can find any other using identities $1+\tan^2(\theta)=\sec^2(\theta) \implies \sec(\theta)=\sqrt{1+\tan^2(\theta)} \implies \cos(\theta)=\frac{1}{\sqrt{1+\tan^2(\theta)}}$

59. kittiwitti1

http://prntscr.com/8dilgn Actually I need help on this first

60. kittiwitti1

LOL sorry

61. kittiwitti1

Mostly because idk what that symbol means

62. anonymous

So which one first?

63. kittiwitti1

The one posted just now

64. kittiwitti1

I only need to know what the symbol is really

65. anonymous

ok $\cos(\theta)=\frac{2\sqrt{13}}{13}$ One ratio you can find right off the start is secant $\sec(\theta)=\frac{1}{\cos(\theta)}=\frac{13}{2\sqrt{13}}$

66. anonymous

$\in$ This symbol?

67. kittiwitti1

Yes, that one. D:

68. anonymous

69. anonymous

70. kittiwitti1

1st? o_o (College)

71. anonymous

The reason they've given this information will be clear after you read this suppose we want to find sin(theta) now we can do $\sin^2(\theta)+\cos^2(\theta)=1$$\sin^2(\theta)=1-\cos^2(\theta)$$\implies |\sin(\theta)|=\sqrt{1-\cos^2(\theta)}$$\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}$ Since theta is in 4th quadrant, sign will be negative for sin(Theta) so we have $\sin(\theta)=-\sqrt{1-\cos^2(\theta)}$

72. anonymous

How come you're studying such basic trigonometry in college? O.o

73. kittiwitti1

Didn't take trig in high school. ...the timer reset at 11:59 and erased half the questions so I got 51%... WTF internet

74. anonymous

Oh, our syllabus is fixed for school, every student that studies mathematics in high school has studied trigonometry and basic calculus here

75. kittiwitti1

http://prntscr.com/8dipej ...I hate this website. Hence why I prefer PHYSICAL tests!

76. anonymous

>_<

77. kittiwitti1

Well since I didn't finish on time apparently (aka stupid website cleared off half my answers) I guess I'll just close the question = =;

78. anonymous