## Mindblast3r one year ago help!

1. mindblast3r

|dw:1436166754136:dw|

2. Compassionate

What do you need help solving

3. mindblast3r

find cosx

4. Compassionate

Do you know the relationship between cosx and sinx?

5. mindblast3r

no

6. LynFran

|dw:1436174086784:dw|

7. imqwerty

cosx = (root8)/3

8. Compassionate

The relationship between the cosine and sine graphs is that the cosine is the same as the sine — only it’s shifted to the left by 90 degrees, or π/2

9. mindblast3r

|dw:1436167048145:dw|

10. mindblast3r

x=2?

11. LynFran

|dw:1436174284089:dw|

12. mindblast3r

i don't even understand what i'm asking to be honest but i have a test on this soon. so pls explain to me what i'm asking.

13. mindblast3r

when i say,

14. mindblast3r

|dw:1436167321407:dw|

15. mindblast3r

what does this mean?

16. mindblast3r

sin x =1/3 Find cos x

17. mindblast3r

the question is that.

18. mindblast3r

i don't get what it's asking me??

19. imqwerty

$cosx = 2\sqrt{2}/3$

20. Astrophysics

Sinx = 1/3 that means the ratio opp/hyp -1/3, so we can make a triangle as following|dw:1436167626215:dw| now we need to figure out cosx, and remember the ratio of cosx is adj/hyp, so solve for the missing length using pythagorean theorem, and you will get your ratio, that make sense?

21. mindblast3r

DOES X MEAN INVERSE?

22. Astrophysics

Huh? No they put x there to represent it's an unkown, and that's what you're solving for

23. Astrophysics

|dw:1436167875535:dw| x is the adjacent side that will help you find cosx

24. mindblast3r

ohh it shows me the answer is |dw:1436167868096:dw|

25. Astrophysics

Well, lets see if its right

26. Astrophysics

We'll use $a^2+b^2 = c^2$ where c is the hypotenuse. We'll let a = 1, b the side we are solving for, and c = 3 $1^2+b^2 = 3^2 \implies b^2= 3^2-1^2 \implies b=\sqrt{3^2-1^2} = \sqrt{9-1} = \sqrt{8}$

27. Astrophysics

So that's our x on the triangle |dw:1436168040178:dw| now we're looking for cosx right, $cosx = \frac{ adj }{ hyp } = \frac{ \sqrt{8} }{ 3 }$

28. Astrophysics

So yes, it simplifies to $\frac{ 2\sqrt{2} }{ 3 }$

29. mindblast3r

$$\color{blue}{\text{Originally Posted by}}$$ @Astrophysics We'll use $a^2+b^2 = c^2$ where c is the hypotenuse. We'll let a = 1, b the side we are solving for, and c = 3 $1^2+b^2 = 3^2 \implies b^2= 3^2-1^2 \implies b=\sqrt{3^2-1^2} = \sqrt{9-1} = \sqrt{8}$ $$\color{blue}{\text{End of Quote}}$$

30. Astrophysics

Something not make sense there?

31. mindblast3r

wait i'm trying to understand.

32. Astrophysics

No worries, take your time

33. anonymous

Yeah, the pythagorean identity in terms of trig functions is written: $\sin^2x+\cos^2x =1$

34. mindblast3r

omg so many people viewing my question.

35. UsukiDoll

@imqwerty please refrain from giving direct answers because it's against OpenStudy's Code of Conduct. "Give Help, Not Answers I will encourage and guide those needing help, and not just give them an answer" "Don’t post only answers - guide the asker to a solution"

36. Astrophysics

Haha, don't get overwhelmed @Mindblast3r just see if you understand everything, if not ask questions

37. UsukiDoll

@Mindblast3r I know how you feel. I had 7 people watching me when I was tutoring sometimes. It was nervewrecking.

38. anonymous

You know, you can really only find $$|\cos x|$$. You need more info to know the sign.

39. mindblast3r

|dw:1436168642801:dw|

40. mindblast3r

come*

41. UsukiDoll

that is from the hypotenuse

42. UsukiDoll

hypotenuse is the longest side of the triangle which was 3 cos x = adjacent / hypotenuse

43. anonymous

$|\cos x| = \sqrt{1 - \sin^2 x}$

44. UsukiDoll

so when you were using the Pythagorean theorem, you were solving for one of the sides of the triangle.

45. Astrophysics

|dw:1436168854387:dw|

46. imqwerty

@UsukiDoll I am sorry :(

47. Astrophysics

What usuki said :), we were just solving for the adjacent side, I should've specified, then we used our cosx ratio adj/hyp

48. Astrophysics

where hyp = 3 and adj = sqrt(8)

49. mindblast3r

|dw:1436168916911:dw|

50. UsukiDoll

@wio 's method is another way of solving this problem, but you should only use it if you know the trig identities...otherwise it's best to use the Pythagorean theorem and drawing a triangle

51. mindblast3r

|dw:1436168957305:dw|

52. Astrophysics

Haha

53. imqwerty

( ͡° ͜ʖ ͡°) u need cosx which is adjacent side/hypotenuse u knw that the hypotenuse is 3 nd nw u need adj. side u solve for adj. side by pythagorus theo. and plug it in the equation nd get answer=(root8)/3

54. UsukiDoll

@imqwerty that's ok...just remember next time that the main reason why OpenStudy exists is to provide guidance towards the question, not a massive online forum to trade answers.

55. UsukiDoll

my spelling and grammar is on vacation xD oy!

56. anonymous

The formula $\sin^2x+\cos^2x=1$Is based on the Pythagorean theorem.

57. Astrophysics

He's not there yet wio lol

58. UsukiDoll

If OP doesn't know those identities or if the problem specifies to use the Pythagorean theorem and draw the triangle method only, so be it.

59. mindblast3r

|dw:1436168987012:dw|

60. Astrophysics

Yes

61. UsukiDoll

yes @Mindblast3r :D

62. imqwerty

so very true

63. mindblast3r

|dw:1436169101210:dw|

64. imqwerty

problem is over nw party time ( ͡^ ͜ʖ ͡^)

65. anonymous

$(\text{adjacent})^2 + (\text{opposite})^2 = (\text{hypotenuse})^2 \\ \iff \frac{(\text{adjacent})^2}{(\text{hypotenuse})^2} + \frac{ (\text{opposite})^2 }{(\text{hypotenuse})^2} = 1 \\ \iff \left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)^2 + \left(\frac{\text{opposite}}{\text{hypotenuse}}\right)^2 = 1 \\ \iff \cos^2x+\sin^2x = 1$

66. mindblast3r

|dw:1436169251275:dw|

67. Astrophysics

Lol, I think usuki, Lyn, and qwerty should get a big thanks to :)

68. mindblast3r

|dw:1436169321629:dw|

69. imqwerty

( ͡° ͜ʖ ͡°)

70. imqwerty

i met a man who said - "i have two daughters and both are girls."

71. mindblast3r

|dw:1436169398594:dw|

72. mindblast3r

ok people shows over, do something else.

73. Astrophysics

Haha ok, bye :). And a way I go... |dw:1436169509148:dw|

74. mindblast3r

lOL!

75. mindblast3r

Now that's funny!