## kaylala one year ago TOPIC: Trigonometry GIVEN THE FOLLOWING CONDITIONS: sin A = 4/5, A Σ I sin B = -3/5, B Σ III Find the value of: I. sin (A +- B) II. cos (A +- B) III. tan (A +- B)

1. kaylala

@Callisto

2. kaylala

@eliassaab

3. eliassaab

I do not understand what comes after 4/5 and -3/5

4. kaylala

let me fix the question

5. eliassaab

Also look at this summary of trig formulas, your problem follows easily http://www.math.missouri.edu/escgi/mucgi-bin/trigidentities.cgi

6. eliassaab

It is still not fixed

7. kaylala

it is. see this pic that i will post. it's where i got the question. wait

8. eliassaab

You need these formulae sin(a + b ) = sin(a) cos(b) + sin(b) cos(a) sin(a - b ) = sin(a) cos(b) - sin(b) cos(a) cos(a + b ) = cos(a) cos(b) - sin(a) sin(b) cos(a - b ) = cos(a) cos(b) + sin(a) sin(b)

9. kaylala

oh it's (+_) the plus on top of the minus for I - III

10. eliassaab

You do first + than -

11. eliassaab

12. eliassaab

Where is the picture?

13. kaylala

wait. i just took a picture of it

14. kaylala

15. Yttrium

@eliassaab , I think the information beside 4/5 and -3/5 are the position of angle in the cartesian plane. (specifically Quadrant I for <A and Quadrant III for <B

16. kaylala

@eliassaab here's the picture

17. kaylala

and yes those are values for the carthesian plane @Yttrium

18. kaylala

i dont think i have to find 6 answers @eliassaab

19. kaylala

20. Yttrium

You really have to find 6 ansers @kaylala. Let me tell you this: $\sin (A \pm B)$ is the notation for$\sin (A+B)$ and $\sin(A-B)$ I hope you understand now.

21. kaylala

oh okay i see now @Yttrium

22. kaylala

i just dont know how to start this

23. Yttrium

24. Yttrium

the condition says sinA = 4/5, where it is in quadrant 1. we know that sinx = opp/hyp Hence, the figure much look like this. |dw:1391508636187:dw|

25. kaylala

okay

26. Yttrium

|dw:1391508787128:dw| let's say that adjacent side of A is x. Can you solve x now?

27. kaylala

okay x=sqrt-9 ??? @Yttrium

28. Yttrium

isn't is sqrt(+9)? which is 3?

29. kaylala

30. kaylala

it do is 3

31. Yttrium

Yea it is. And, with that you can now actually solve for cosA, tanA.

32. kaylala

how?

33. kaylala

i really dont know how to do this please guide me @Yttrium

34. Yttrium

35. kaylala

sorta

36. kaylala

cos A = 3/5 ???

37. kaylala

tanA=4/3

38. kaylala

did i get it?

39. kaylala

what do i do next?

40. Yttrium

You got this. I think it is better if you write sinA, cosA, and tanA in a piece of paper. They are really important later. Next thing we are going to do is to solve for cosB and tanB

41. kaylala

yeah. just wrote it down already how do i solve for that?

42. kaylala

y=4

43. kaylala

? @Yttrium

44. Yttrium

you sure? isn't it -4? take note it is in the quadrant III.

45. kaylala

okay so the quadrant determines the sign? right?

46. Yttrium

Yeah. for (x,y) quadrant I (+,+) II (-,+) III (-,-) IV (+,-) Now. Find cosB and tanB. Be careful with the signs.

47. kaylala

okay then

48. Yttrium

Solve for cosB and tanB

49. kaylala

cos B= -4/5

50. kaylala

tanB= 3/-4

51. Yttrium

Ohhh wait!!!

52. kaylala

did i get it?

53. kaylala

ookay

54. Yttrium

the graph will be like this |dw:1391510194742:dw|

55. Yttrium

x then is -4 solve again for cosB and tanB

56. kaylala

i see now but the answer for cos b and tan B still remains?

57. kaylala

58. kaylala

what to do next?

59. kaylala

@Yttrium

60. Yttrium

ooops tanB = -3/-4 = 3/4

61. kaylala

oh you're right

62. Yttrium

Now is the time to evaluate them. I. sin (A +- B) = sinAcosB +- cosAsinB II. cos (A +- B) = cosAcos -+ sinAsinB

63. Yttrium

III. $\tan(A \pm B) = \frac{ tanA \pm tanB }{ 1\mp tanAtanB }$

64. kaylala

how's that? do i just substitute what i got? to the given formulas?

65. Yttrium

Yeah. All the operation found upside are the only operations used when applying upside identities. Look at this. $\tan(A +B) = \frac{ tanA + tanB }{ 1 - tanAtanB }$

66. kaylala

so sin(a+b) = 1 sin(a-b) = -7/25 ??? @Yttrium

67. kaylala

cos (a+b)=0 cos (a-b)=-24/25 did i get it correctly @Yttrium ?

68. Yttrium

i'll just check the sin function, you take care of the others :D sin(a+b) = -1

69. kaylala

okay :) thanks

70. kaylala

tan (A+B)= undefined??? is it? please check this @Yttrium

71. Yttrium

what is tanA and tanbB?

72. kaylala

tanA=4/3 tanB=3/4

73. Yttrium

Yeah. That's undefined. And that is normal. :))

74. kaylala

i see. okay thanks again :)

75. Yttrium

No problem :))