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SugarRainbow Group Title

find the value of the expression if sin beta= 0.45, find cos(pi/2-beta)

  • 2 years ago
  • 2 years ago

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  1. SugarRainbow Group Title
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    do i just do pi/2-0.45?

    • 2 years ago
  2. lgbasallote Group Title
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    sin beta is 0.45...but beta is not 0.45 does that make sense?

    • 2 years ago
  3. lgbasallote Group Title
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    \[\Large \cos (\frac \pi 2 - \beta) \implies \cos \frac \pi 2 \cos \beta + \sin \frac \pi 2 \sin \beta\] are you familiar with this?

    • 2 years ago
  4. SugarRainbow Group Title
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    no idk what you just did

    • 2 years ago
  5. lgbasallote Group Title
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    it's a trig formula \[\LARGE \cos (A - B ) \implies \cos A \cos B + \sin A \sin B\]

    • 2 years ago
  6. lgbasallote Group Title
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    in this case, our A is pi/2 and B is beta make sense?

    • 2 years ago
  7. SugarRainbow Group Title
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    yeah but whats beta?

    • 2 years ago
  8. Algebraic! Group Title
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    |dw:1347254638009:dw|

    • 2 years ago
  9. lgbasallote Group Title
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    ahh we'll get there

    • 2 years ago
  10. Algebraic! Group Title
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    already did.

    • 2 years ago
  11. lgbasallote Group Title
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    so you get how \[\cos(\frac \pi 2 - \beta) = \cos \frac \pi 2 \cos \beta + \sin \frac \pi 2 \sin \beta\]

    • 2 years ago
  12. Algebraic! Group Title
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    answer is clearly .45

    • 2 years ago
  13. lgbasallote Group Title
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    i was talking to sugarrainbow

    • 2 years ago
  14. Skaematik Group Title
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    Is it just not cos(pi/2 - sin^-1(0.45))?

    • 2 years ago
  15. SugarRainbow Group Title
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    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

    • 2 years ago
  16. lgbasallote Group Title
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    it works because \[\cos \frac \pi 2 = 0\] \[\sin \frac \pi 2 = 1\]

    • 2 years ago
  17. lgbasallote Group Title
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    so if you substitute you get \[\implies (0) \cos \beta + (1) \sin \beta\]

    • 2 years ago
  18. SugarRainbow Group Title
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    what about what Skaematik did? can't i just do that?

    • 2 years ago
  19. SugarRainbow Group Title
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    and how come we add cos+sin?

    • 2 years ago
  20. Skaematik Group Title
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    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

    • 2 years ago
  21. tanjung Group Title
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    cos(pi/2-beta) = sin beta

    • 2 years ago
  22. SugarRainbow Group Title
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    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?

    • 2 years ago
  23. SugarRainbow Group Title
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    when i did pi/2 i got 0.5

    • 2 years ago
  24. tanjung Group Title
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    tangent (pi/2-beta) still in the first kuadrant, so impossible has a negative value and tangent (pi/2-beta)=cotangent beta

    • 2 years ago
  25. SugarRainbow Group Title
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    okay so is that what cosine is or how do we get that

    • 2 years ago
  26. tanjung Group Title
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    maybe ur problem is if given tangent(pi/2-beta)=5.32, find cot beta? cot beta = tan(pi/2-beta) = 5.32

    • 2 years ago
  27. Algebraic! Group Title
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    |dw:1347256883382:dw|

    • 2 years ago
  28. Algebraic! Group Title
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    cotangent of beta = 1/-5.32

    • 2 years ago
  29. SugarRainbow Group Title
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    how did you get -5.32

    • 2 years ago
  30. Algebraic! Group Title
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    see beta?

    • 2 years ago
  31. Algebraic! Group Title
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    what's the tangent? opp = -5.32 adjacent =1

    • 2 years ago
  32. Algebraic! Group Title
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    tan(beta) = -5.32 and cot(beta) = 1/-5.32

    • 2 years ago
  33. SugarRainbow Group Title
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    but how did you get that if we have sin and cos?

    • 2 years ago
  34. Algebraic! Group Title
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    what problem are you working on?

    • 2 years ago
  35. SugarRainbow Group Title
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    oh i'm sorry i was still looking at the first one

    • 2 years ago
  36. SugarRainbow Group Title
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    can we just go back to the first one cuz i still don't understand it

    • 2 years ago
  37. Algebraic! Group Title
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    sure, look carefully at the sketch plz

    • 2 years ago
  38. Algebraic! Group Title
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    |dw:1347257483078:dw|

    • 2 years ago
  39. Algebraic! Group Title
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    do you agree? sin(beta) = .45/1

    • 2 years ago
  40. Algebraic! Group Title
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    |dw:1347257532660:dw|

    • 2 years ago
  41. Algebraic! Group Title
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    opposite over hypotenuse is sine

    • 2 years ago
  42. SugarRainbow Group Title
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    okay that makes sense

    • 2 years ago
  43. Algebraic! Group Title
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    cool:)

    • 2 years ago
  44. SugarRainbow Group Title
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    but now how do i get cos?

    • 2 years ago
  45. Algebraic! Group Title
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    now: |dw:1347257619039:dw| what's that new angle?

    • 2 years ago
  46. Algebraic! Group Title
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    the whole angle from the x axis to the y axis is 90 or pi/2 ....

    • 2 years ago
  47. Algebraic! Group Title
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    so that angle must be pi/2 - Beta

    • 2 years ago
  48. SugarRainbow Group Title
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    does this have anything to do with the unit circle

    • 2 years ago
  49. Algebraic! Group Title
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    now look at the cosine of that angle: |dw:1347257708887:dw|

    • 2 years ago
  50. Algebraic! Group Title
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    yeah, it all is the same thing..

    • 2 years ago
  51. Algebraic! Group Title
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    see that triangle. one side is .45 the hypotenuse is 1 the other side, no one cares what it is!

    • 2 years ago
  52. Algebraic! Group Title
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    |dw:1347257807464:dw|

    • 2 years ago
  53. Algebraic! Group Title
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    the cosine of that new angle is clearly .45 /1 .45 is adjacent, 1 is the hypotenuse.

    • 2 years ago
  54. Algebraic! Group Title
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    so cos(pi/2 -Beta) is .45

    • 2 years ago
  55. Algebraic! Group Title
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    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.

    • 2 years ago
  56. Algebraic! Group Title
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    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...

    • 2 years ago
  57. Algebraic! Group Title
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    |dw:1347258175266:dw|

    • 2 years ago
  58. Algebraic! Group Title
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    |dw:1347258227158:dw|

    • 2 years ago
  59. Algebraic! Group Title
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    |dw:1347258316959:dw|

    • 2 years ago
  60. SugarRainbow Group Title
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    umm okay is it also because its sin and cos that its the same?

    • 2 years ago
  61. SugarRainbow Group Title
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    real quick, since pi/2=1 then would i just divide .45(sine beta) by 1 to get cos?

    • 2 years ago
  62. Algebraic! Group Title
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    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

    • 2 years ago
  63. Algebraic! Group Title
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    pi/2 != 1

    • 2 years ago
  64. Algebraic! Group Title
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    pi/2 = pi/2

    • 2 years ago
  65. SugarRainbow Group Title
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    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|

    • 2 years ago
  66. Algebraic! Group Title
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    ah the hypotenuse =1 you mean?

    • 2 years ago
  67. SugarRainbow Group Title
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    yeah

    • 2 years ago
  68. Algebraic! Group Title
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    cosine isn't the same as sine.... that's not what I said, but yeah you're starting to get it, I think...

    • 2 years ago
  69. Algebraic! Group Title
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    your pic is accurate

    • 2 years ago
  70. Algebraic! Group Title
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    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:)

    • 2 years ago
  71. SugarRainbow Group Title
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    oh so its only the same cuz it the angle adds up to 90?

    • 2 years ago
  72. Algebraic! Group Title
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    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 :)

    • 2 years ago
  73. SugarRainbow Group Title
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    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?

    • 2 years ago
  74. Algebraic! Group Title
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    well remember, because if Beta is one of the angles and the other is pi/2 - Beta then Beta + (pi/2 - Beta) = pi/2

    • 2 years ago
  75. SugarRainbow Group Title
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    but what's pi/2?

    • 2 years ago
  76. Algebraic! Group Title
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    so we're automatically talking about two angles that add up to pi/2.

    • 2 years ago
  77. Algebraic! Group Title
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    pi/2 is 90 deg.

    • 2 years ago
  78. Algebraic! Group Title
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    1/4 the way around a circle.

    • 2 years ago
  79. SugarRainbow Group Title
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    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

    • 2 years ago
  80. Algebraic! Group Title
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    that's good actually. that means it works no matter what Beta is. It's general. For any Beta.

    • 2 years ago
  81. Algebraic! Group Title
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    |dw:1347259786709:dw|

    • 2 years ago
  82. Algebraic! Group Title
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    |dw:1347259882928:dw|

    • 2 years ago
  83. SugarRainbow Group Title
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    okay

    • 2 years ago
  84. Algebraic! Group Title
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    still talking about first quadrant stuff at the moment..

    • 2 years ago
  85. Algebraic! Group Title
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    anyway hope that cleared it up. think it over a bit, you'll see why it has to be true.

    • 2 years ago
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