## hannahhaack one year ago Find the values of the six trigonometric functions of theta: cos θ = 35/37 constraint: tan θ < 0

1. hannahhaack

|dw:1401422933563:dw|

2. dan815

six trignometric functions?

3. hannahhaack

yes. sin, tan, sec, cos, cot, csc

4. dan815

ok plug into calculator using the inverse buton

5. hannahhaack

to find all six? Can you baby it down for me haha

6. dan815

okay lol

7. dan815

so first look at your triangle its 35 in x and -37 in y

8. dan815

so enter 35/-37 in your calculator, and then do cos-1 (35/-37)

9. dan815

thats a button on your calculator cos inverse

10. dan815

do the same for sin and tan

11. hannahhaack

cos=2.811295299?

12. dan815

make sure you are using degree mode

13. hannahhaack

ok it is now cos=161.0753556

14. dan815

15. dan815

alright thats one, now do the rest

16. dan815

remember that 1/sin= csc 1/cos=sec 1/tan=cot

17. dan815

by the way that is not the only solution,

18. hannahhaack

ya it usually accepts fractions.

19. dan815

|dw:1401423483117:dw|

20. dan815

it found that angle, but if your teacher is particular on you to only find tht last angle on the bottom

21. hannahhaack

So i just plugged in -37/$\sqrt{2594}$

22. hannahhaack

is it in quadrant 2 not 4?

23. dan815

i dont know its your choice what do you want??

24. hannahhaack

well the cos is positive and the tan is negative. so wont that be quad 4?

25. dan815

yeah

26. hannahhaack

ok now what would the hypotenus equal?

27. dan815

do Pythagorean thm

28. dan815

|dw:1401423730828:dw|

29. dan815

ok

30. hannahhaack

i get square root of 2594

31. hannahhaack

is that right? I might be doing my calcs wrong

32. dan815

33. myininaya

|dw:1401423880441:dw|

34. hannahhaack

so would sin of theta=37/square root of 2594

35. dan815

the hypotenuse should be roughly sqrt2*36

36. hannahhaack

it says it is wrong

37. hannahhaack

square root of 36?

38. dan815

its -37

39. dan815

no nvm that, its just an approximation i dont have calc

40. myininaya

cos is adj/hyp you labeled the initial triangle a bit wrong

41. dan815

okay do it over

42. hannahhaack

can you show me the right triangle?

43. dan815

|dw:1401424044893:dw|

44. myininaya

good

45. myininaya

now find the opp measurement of the angle theta |dw:1401424168943:dw|

46. hannahhaack

and how did you get that?

47. myininaya

use the theorem we talked about last question

48. myininaya

you are given cos is 35/37

49. hannahhaack

right right right. got it

50. hannahhaack

none of those are negative?

51. myininaya

right cos is positive not negative

52. myininaya

sin is negative

53. myininaya

so you will definitely get a negative number for that opp measurement to theta

54. hannahhaack

i got 12

55. myininaya

and it should really be -12

56. hannahhaack

35^2+b^2=37^2

57. hannahhaack

b^2=144 b=12?

58. myininaya

or b=-12 we will go with b=-12 because we were given cos is + tan is - which makes sin -

59. hannahhaack

ok. so sin =-12/37

60. myininaya

|dw:1401424409906:dw| that symbolizes the triangle is to the right of zero but below 0 (that is talking about the signs)

61. myininaya

and yes

62. hannahhaack

that clarifies completely. what about this one? tan θ is undefined. constraint: π ≤ θ ≤ 2π

63. hannahhaack

find the six.

64. dan815

remember tan theta = sin theta/costheta so its undefin when cos theta= 0 which is at?

65. hannahhaack

so each one would be undefined?

66. dan815

no just the tan and the cot

67. dan815

well atually no cot is fine

68. hannahhaack

how do i find the rest?

69. dan815

have to first find this

70. dan815

tan theta= sin theta/cos theta => undefined when denominator =0 where cos theta =0

71. dan815

solve for cos theta = 0 , when theta is in that given interval pi to 2pi

72. hannahhaack

would it be 2pi/pi?

73. dan815

no