## SugarRainbow Group Title find the value of the expression if sin beta= 0.45, find cos(pi/2-beta) one year ago one year ago

1. SugarRainbow Group Title

do i just do pi/2-0.45?

2. lgbasallote Group Title

sin beta is 0.45...but beta is not 0.45 does that make sense?

3. lgbasallote Group Title

$\Large \cos (\frac \pi 2 - \beta) \implies \cos \frac \pi 2 \cos \beta + \sin \frac \pi 2 \sin \beta$ are you familiar with this?

4. SugarRainbow Group Title

no idk what you just did

5. lgbasallote Group Title

it's a trig formula $\LARGE \cos (A - B ) \implies \cos A \cos B + \sin A \sin B$

6. lgbasallote Group Title

in this case, our A is pi/2 and B is beta make sense?

7. SugarRainbow Group Title

yeah but whats beta?

8. Algebraic! Group Title

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9. lgbasallote Group Title

ahh we'll get there

10. Algebraic! Group Title

11. lgbasallote Group Title

so you get how $\cos(\frac \pi 2 - \beta) = \cos \frac \pi 2 \cos \beta + \sin \frac \pi 2 \sin \beta$

12. Algebraic! Group Title

13. lgbasallote Group Title

i was talking to sugarrainbow

14. Skaematik Group Title

Is it just not cos(pi/2 - sin^-1(0.45))?

15. SugarRainbow Group Title

umm kay i don't get algebraic's picture and i not sure how igbasallote's formula works if we don't know what beta is

16. lgbasallote Group Title

it works because $\cos \frac \pi 2 = 0$ $\sin \frac \pi 2 = 1$

17. lgbasallote Group Title

so if you substitute you get $\implies (0) \cos \beta + (1) \sin \beta$

18. SugarRainbow Group Title

what about what Skaematik did? can't i just do that?

19. SugarRainbow Group Title

and how come we add cos+sin?

20. Skaematik Group Title

This is a formula that you will need to memorise. Its something that you can use to manipulate trignometric equations to find an answer cos(A−B)⟹cosAcosB+sinAsinB Therefore if you use it in this case, like lgbasallote said cos(π2−β)=cosπ/2cosβ+sinπ/2sinβ Plug these values into your calculator cosπ/2 sinπ/2 And you will find they equal to 0 and 1 Therefore cos(π2−β)=0+1*sinβ Sinβ =0.45 Therefore the answer is 0.45

21. tanjung Group Title

cos(pi/2-beta) = sin beta

22. SugarRainbow Group Title

okay that makes more sense but what would the formula be if i had tangent(pi/2-beta)=-5.32 and i have to find cot?

23. SugarRainbow Group Title

when i did pi/2 i got 0.5

24. tanjung Group Title

tangent (pi/2-beta) still in the first kuadrant, so impossible has a negative value and tangent (pi/2-beta)=cotangent beta

25. SugarRainbow Group Title

okay so is that what cosine is or how do we get that

26. tanjung Group Title

maybe ur problem is if given tangent(pi/2-beta)=5.32, find cot beta? cot beta = tan(pi/2-beta) = 5.32

27. Algebraic! Group Title

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28. Algebraic! Group Title

cotangent of beta = 1/-5.32

29. SugarRainbow Group Title

how did you get -5.32

30. Algebraic! Group Title

see beta?

31. Algebraic! Group Title

what's the tangent? opp = -5.32 adjacent =1

32. Algebraic! Group Title

tan(beta) = -5.32 and cot(beta) = 1/-5.32

33. SugarRainbow Group Title

but how did you get that if we have sin and cos?

34. Algebraic! Group Title

what problem are you working on?

35. SugarRainbow Group Title

oh i'm sorry i was still looking at the first one

36. SugarRainbow Group Title

can we just go back to the first one cuz i still don't understand it

37. Algebraic! Group Title

sure, look carefully at the sketch plz

38. Algebraic! Group Title

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39. Algebraic! Group Title

do you agree? sin(beta) = .45/1

40. Algebraic! Group Title

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41. Algebraic! Group Title

opposite over hypotenuse is sine

42. SugarRainbow Group Title

okay that makes sense

43. Algebraic! Group Title

cool:)

44. SugarRainbow Group Title

but now how do i get cos?

45. Algebraic! Group Title

now: |dw:1347257619039:dw| what's that new angle?

46. Algebraic! Group Title

the whole angle from the x axis to the y axis is 90 or pi/2 ....

47. Algebraic! Group Title

so that angle must be pi/2 - Beta

48. SugarRainbow Group Title

does this have anything to do with the unit circle

49. Algebraic! Group Title

now look at the cosine of that angle: |dw:1347257708887:dw|

50. Algebraic! Group Title

yeah, it all is the same thing..

51. Algebraic! Group Title

see that triangle. one side is .45 the hypotenuse is 1 the other side, no one cares what it is!

52. Algebraic! Group Title

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53. Algebraic! Group Title

the cosine of that new angle is clearly .45 /1 .45 is adjacent, 1 is the hypotenuse.

54. Algebraic! Group Title

so cos(pi/2 -Beta) is .45

55. Algebraic! Group Title

it's what @tanjung said : cos(pi/2-beta) = sin beta but I showed it graphical so you could believe! you can't really argue with that picture.

56. Algebraic! Group Title

basically if two angles are complimentary, the cosine of one angle is the sin of the other.... ie the side that is opposite on one triangle *must be* the side that is adjacent on the other triangle...

57. Algebraic! Group Title

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58. Algebraic! Group Title

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59. Algebraic! Group Title

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60. SugarRainbow Group Title

umm okay is it also because its sin and cos that its the same?

61. SugarRainbow Group Title

real quick, since pi/2=1 then would i just divide .45(sine beta) by 1 to get cos?

62. Algebraic! Group Title

yeah, I guess I get what you're saying... it comes from the definition of sin as opp/hyp and cosine as adj/hyp when you draw those sides for two angles which are complimentary they make a rectangle

63. Algebraic! Group Title

pi/2 != 1

64. Algebraic! Group Title

pi/2 = pi/2

65. SugarRainbow Group Title

umm okay so thinking of it in terms of a rectangle that means cos is the same as sin because the two sides are equal?|dw:1347258695721:dw|

66. Algebraic! Group Title

ah the hypotenuse =1 you mean?

67. SugarRainbow Group Title

yeah

68. Algebraic! Group Title

cosine isn't the same as sine.... that's not what I said, but yeah you're starting to get it, I think...

69. Algebraic! Group Title

70. Algebraic! Group Title

you see that, in your pic: |dw:1347258903616:dw| those two angles aren't the same? they do add up to give pi/2 (90 degrees) though so the cosine of one angle = the sine of the other:)

71. SugarRainbow Group Title

oh so its only the same cuz it the angle adds up to 90?

72. Algebraic! Group Title

no, it works elsewhere on the circle. but just try to grasp this 1st quadrant sketch for now... then you can move on to more advanced stuff :)

73. SugarRainbow Group Title

umm okay i do get how you got hyp=1 and opp=.45 but what bout the angles how did you get that they'd = 90?

74. Algebraic! Group Title

well remember, because if Beta is one of the angles and the other is pi/2 - Beta then Beta + (pi/2 - Beta) = pi/2

75. SugarRainbow Group Title

but what's pi/2?

76. Algebraic! Group Title

so we're automatically talking about two angles that add up to pi/2.

77. Algebraic! Group Title

pi/2 is 90 deg.

78. Algebraic! Group Title

1/4 the way around a circle.

79. SugarRainbow Group Title

oh i just put it in my calc and i see how you got 90 but how does that necessarily help us with this because a bunch of different combinations of angles can add up to 90

80. Algebraic! Group Title

that's good actually. that means it works no matter what Beta is. It's general. For any Beta.

81. Algebraic! Group Title

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82. Algebraic! Group Title

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83. SugarRainbow Group Title

okay

84. Algebraic! Group Title