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## lilsis76 3 years ago 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

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1. lilsis76

|dw:1350692195063:dw|

2. surdawi

|dw:1350692395385:dw|

3. lilsis76

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

4. surdawi

csc B = 1/sin B

5. lilsis76

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

6. surdawi

yes

7. lilsis76

|dw:1350692578128:dw| right?

8. surdawi

yes, next

9. lilsis76

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

10. lilsis76

im trying to find the identities

11. surdawi

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

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

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

14. lilsis76

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

15. surdawi

what does x = ?

16. surdawi

|dw:1350693950387:dw|

17. surdawi

by the way what grade is this

18. lilsis76

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

19. lilsis76

x = sqrt 13

20. surdawi

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

21. lilsis76

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

22. lilsis76

okay so the cos=3sqrt13 /26

23. surdawi

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

24. surdawi

can you calculate theta now

25. lilsis76

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

26. lilsis76

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

27. lilsis76

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

thats the cos of thata, dont you want theta?

29. lilsis76

yes the theta of tangent

30. lilsis76

wait. cos theta = the 3sqrt13 / 13

31. lilsis76

im so lost

32. surdawi

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

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

34. Outkast3r09

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

35. Outkast3r09

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

36. lilsis76

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

37. Outkast3r09

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

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

39. Outkast3r09

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

40. lilsis76

|dw:1350697001280:dw|

41. Outkast3r09

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

42. Outkast3r09

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

45. Outkast3r09

how would you get theta by itself

46. Outkast3r09

do you know about inverse trig functions?

47. lilsis76

im back server was werid. let me look

48. lilsis76

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

49. Outkast3r09

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

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

51. Outkast3r09

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

52. lilsis76

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

53. Outkast3r09

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

54. lilsis76

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

55. Outkast3r09

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

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

57. Outkast3r09

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

58. lilsis76

so the tan^-1 is called arctan?

59. Outkast3r09

yes =]

60. lilsis76

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

alright where is -33.69 degrees?

62. lilsis76

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

63. Outkast3r09

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

OH, umm... well hold on

65. lilsis76

|dw:1350698454198:dw|

66. Outkast3r09

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

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

68. Outkast3r09

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

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

70. lilsis76

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

71. lilsis76

i cant find the equ.

72. lilsis76

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

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

74. jim_thompson5910

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

75. lilsis76

yes

76. jim_thompson5910

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

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

78. jim_thompson5910

so sin(theta) = 24/25

79. lilsis76

lol let me check

80. jim_thompson5910

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

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

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

yeah i fixed that error, my bad

84. lilsis76

lol just a moment

85. lilsis76

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

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

yeah i fixed that

88. lilsis76

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

89. lilsis76

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

yw

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