## lilsis76 Group Title Given tan theta = -2/3, where pi/2 < 0 < pi and csc B=2 where 0<B<pi/2 Find sin (0+B) and cos (B-0) ill draw the 3 data thingys given one year ago one year ago

1. lilsis76 Group Title

|dw:1350692195063:dw|

2. surdawi Group Title

|dw:1350692395385:dw|

3. lilsis76 Group Title

based on the information i know it will be first quad.

4. surdawi Group Title

csc B = 1/sin B

5. lilsis76 Group Title

oh duh haha its negative, so its going backwards right?

6. surdawi Group Title

yes

7. lilsis76 Group Title

|dw:1350692578128:dw| right?

8. surdawi Group Title

yes, next

9. lilsis76 Group Title

dang it, im going to need identities right? cuz it kinda looks like it :/

10. lilsis76 Group Title

im trying to find the identities

11. surdawi Group Title

1st find theta and beta sin (0 + B) = sin 0 cos B + cos 0 sin B cos (0 – B) = cos 0 cos B + sin 0 sin B

12. lilsis76 Group Title

UGH!!! okay im back lol okay, are these the identities I should know for finding theta and beta then? just so i save it?

13. surdawi Group Title

$\tan \theta=\frac{ -2 }{ 3 }$ $\tan^{-1} (\tan \theta)=\tan^{-1} (\frac{ -2 }{ 3 })$

14. lilsis76 Group Title

okay so far i know: sin = 2/x cos =2/x and tan is the same -2/3

15. surdawi Group Title

what does x = ?

16. surdawi Group Title

|dw:1350693950387:dw|

17. surdawi Group Title

by the way what grade is this

18. lilsis76 Group Title

okay hold on, this is precal. im in a community college, but im horrible at math haha

19. lilsis76 Group Title

x = sqrt 13

20. surdawi Group Title

been a while since i been in the class anyways x is correct

21. lilsis76 Group Title

oh really? did u graduate? and YAY okay i got the x, now will that be the theta?

22. lilsis76 Group Title

okay so the cos=3sqrt13 /26

23. surdawi Group Title

right now im working on my|dw:1350694287644:dw|

24. surdawi Group Title

can you calculate theta now

25. lilsis76 Group Title

uh.... well the one i had or placed on here was wrong? the okay so the cos=3sqrt13 /26

26. lilsis76 Group Title

is the 2 a regular 2, or a negative 2? since its going downward of the unit circle?

27. lilsis76 Group Title

wait im lost: i want to go back a lil to the soh cah toa cah is adj/hyp that would be 3 / sqrt13 |dw:1350694753458:dw| right?

28. surdawi Group Title

thats the cos of thata, dont you want theta?

29. lilsis76 Group Title

yes the theta of tangent

30. lilsis76 Group Title

wait. cos theta = the 3sqrt13 / 13

31. lilsis76 Group Title

im so lost

32. surdawi Group Title

can i do it on a piece of paper then send it to you, im tired and want to go to sleep; yes i know its only 9:20 lol

33. lilsis76 Group Title

lol yes you may. and thanks, im getting a headache from this too

34. Outkast3r09 Group Title

alright so first we need to figure out where tan is negative. Using this circle |dw:1350696636481:dw|

35. Outkast3r09 Group Title

these are when the functions are positive. for instance, in quadrant 1 , all function are positive

36. lilsis76 Group Title

okay so tan would be negative in...................the cosine area

37. Outkast3r09 Group Title

and also the sine , because tangent is $\frac{sin\theta}{cos\theta}$... so anytime either one is negative tan will be negative ... when sin and cos are both negative ,tan is positive which is this quadrant|dw:1350696881960:dw|

38. lilsis76 Group Title

okay so the 2nd and 3rd piece on the left sides

39. Outkast3r09 Group Title

next we use the information it gives about theta $\pi/2<\theta<\pi$ can you mark these angles on my picture

40. lilsis76 Group Title

|dw:1350697001280:dw|

41. Outkast3r09 Group Title

so theta is less than $\pi/2$ but greater than $\pi$|dw:1350697010218:dw|

42. Outkast3r09 Group Title

this means $\theta is in your second quadrant 43. lilsis76 Group Title okay i see that now 44. Outkast3r09 Group Title alright so now lets solve the equation \[tan(\theta)=\frac{-2}{3}$

45. Outkast3r09 Group Title

how would you get theta by itself

46. Outkast3r09 Group Title

do you know about inverse trig functions?

47. lilsis76 Group Title

im back server was werid. let me look

48. lilsis76 Group Title

no i dont know about the inverse trig functions unless your talkin about it being 1/tan or like the cos/sin

49. Outkast3r09 Group Title

inverse function work as "cancelers" of trig functions they're usually denoted as $arcsin,arctan,arccos$ or $sin^{-1},cos^{-1}$ you know them because you know 1/tanx. I think you just didn't know what they were called

50. Outkast3r09 Group Title

these cancel out the same trig function so $arcsin(sin(x))=x$

51. Outkast3r09 Group Title

so what would you do in your case $tan(\theta)=\frac{-2}{3}$

52. lilsis76 Group Title

no we have not gone over this :/ and so the arctan one would be arctan(tan(x))= -2/3 ?

53. Outkast3r09 Group Title

if you do arctan on one side you have to do it on the other =]

54. lilsis76 Group Title

-2/3arctan(tan(x))=-2/3

55. Outkast3r09 Group Title

ok say $tan(x)=\frac{-2}{3}$ you know to get rid of tan you have to use arctan, however if you multiply the left side by arctan, you have to multiply the right side also $arctan(tan(x)=arctan(\frac{-2}{3})$

56. Outkast3r09 Group Title

since arctan cancels tan you get $x=arctan(\frac{-2}{3})$

57. Outkast3r09 Group Title

your calculator should have an arctan symbol. it might be $tan^{-1}$

58. lilsis76 Group Title

so the tan^-1 is called arctan?

59. Outkast3r09 Group Title

yes =]

60. lilsis76 Group Title

weird....okay, ya i didnt know any inverse ones i couldnt find the identities either :/ but i did tan^-1(-2/3) and it gave me.............-33.69

61. Outkast3r09 Group Title

alright where is -33.69 degrees?

62. lilsis76 Group Title

haha you really ask me degreees?? umm... i think its that 180/pi, right?

63. Outkast3r09 Group Title

theta is in degrees unless you put in radians but keep it in degrees , all i'm asking is where on a circle would -33.69 degrees be( quadrant wise)

64. lilsis76 Group Title

OH, umm... well hold on

65. lilsis76 Group Title

|dw:1350698454198:dw|

66. Outkast3r09 Group Title

well uh oh that isn't in the second quadrant we need to find an angle that is in the second quadrant with the same value as -33.69 degrees

67. lilsis76 Group Title

should i start from 90 degrees?? |dw:1350698677314:dw|

68. Outkast3r09 Group Title

hold on i know what you need to do just don't know why it's been a while since going through this material

69. Outkast3r09 Group Title

i know you have to add pi(180 degrees) to it

70. lilsis76 Group Title

okay, i think its mult. the ( 180/pi) i think

71. lilsis76 Group Title

i cant find the equ.

72. lilsis76 Group Title

WAIT!!! to change to degrees you multiply 180 deg/ pi. but i would need the point for the other one to do it..... here is an example from my notes: pi/12(180/pi) cross reduce to--->180 deg./12 = 15 degrees---sin15degrees =.2588

73. lilsis76 Group Title

I really cant figure this one out... can you? or are you writingit down on paper, or looking it up online??

74. jim_thompson5910 Group Title

in your drawing at the very top, you say cos 7/25 did you mean cos(theta) = 7/25 ???

75. lilsis76 Group Title

yes

76. jim_thompson5910 Group Title

so because sin^2(x) + cos^2(x) = 1 you can find sin(theta) like so sin^2(theta) + cos^2(theta) = 1 sin^2(theta) + (7/25)^2 = 1 sin^2(theta) + 49/625 = 1 sin^2(theta) = 1 - 49/625 sin^2(theta) = 576/625 sin(theta) = -sqrt(576/625) sin(theta) = -24/25

77. jim_thompson5910 Group Title

oh wait, sry sine is positive because 0 < theta < pi/2

78. jim_thompson5910 Group Title

so sin(theta) = 24/25

79. lilsis76 Group Title

lol let me check

80. jim_thompson5910 Group Title

sin^2(B) + cos^2(B) = 1 sin^2(B) + (-3/5)^2 = 1 sin^2(B) + 9/25 = 1 sin^2(B) = 1 - 9/25 sin^2(B) = 16/25 sin(B) = -sqrt(16/25) ... sin(B) is negative for pi < B < 3pi/2 sin(B) = -4/5

81. jim_thompson5910 Group Title

So after those two steps, we now know these four pieces of info sin(theta) = 24/25 cos(theta) = 7/25 sin(B) = -4/5 cos(B) = -3/25 You can now use them in the identities sin(theta + B) = sin(theta)cos(B) + cos(theta)sin(B) cos(theta - B) = cos(theta)cos(B) + sin(theta)sin(B)

82. lilsis76 Group Title

wait sin B is negative?? because..... its in the ..3rd quardinate and sin is negative and 24/25 is positive because it falls between the |dw:1350700800280:dw| 0 < theta < pi/2

83. jim_thompson5910 Group Title

yeah i fixed that error, my bad

84. lilsis76 Group Title

lol just a moment

85. lilsis76 Group Title

wait sin B is negative?? because..... its in the ..3rd quardinate and sin is negative and 24/25 is positive because it falls between the |dw:1350700800280:dw| 0 < theta < pi/2

86. lilsis76 Group Title

Sin B is negative??? because its in the 3rd quar. and that sin is negative and 24/25 is positive because it falls between the 0 < theta < pi/2

87. jim_thompson5910 Group Title

yeah i fixed that

88. lilsis76 Group Title

okay, im checking my notes cuz i have writing in every direction

89. lilsis76 Group Title

okay, thank you, lol i have to write it neater and go over it again to get things down. thank you for helping me. ill give u a medal

90. jim_thompson5910 Group Title

yw