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anonymous

  • 5 years ago

Verify the identity tan(θ)cot(θ)-sin^2θ=cos^2θ

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  1. bahrom7893
    • 5 years ago
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    Hey, I don't feel like copying and pasting, so I'll just use t instead of theta. tan(t)cot(t)-Sin^2t = Cos^2t. I rewrote it in terms of t, so that it;s easier for me to type

  2. bahrom7893
    • 5 years ago
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    Bear in mind: tan(t) = Sin(t)/Cos(t) and Cot(t) = Cos(t)/Sin(t)

  3. bahrom7893
    • 5 years ago
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    insert those into your equation: [Sin(t)/Cos(t)]*[Cos(t)/Sin(t)] - Sin^2t = Cos^2t Simplify: 1 - Sin^2t = Cos^2t

  4. bahrom7893
    • 5 years ago
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    Well actually Tan*Cot = 1, but I showed you where that one came from.

  5. bahrom7893
    • 5 years ago
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    So now remember the identity: Sin^2 + Cos^2 = 1, well that's the case; move -Sin^2t to the right: 1 = Cos^2t + Sin^2t 1 = 1 <identity is verified

  6. anonymous
    • 5 years ago
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    Ok so I basically am just plugging in what they equal for example tan(t)=sin(t)/cos(t). what if I have a problem with csc... how does that work or what is that equivalent to?

  7. anonymous
    • 5 years ago
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    Ok so I basically am just plugging in what they equal for example tan(t)=sin(t)/cos(t). what if I have a problem with csc... how does that work or what is that equivalent to?

  8. anonymous
    • 5 years ago
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    The best thing to do when proving identities is generally to break everything down in terms of sine and cosine. If you know the identities and break everything down, it's just algebra with trig functions.

  9. anonymous
    • 5 years ago
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    ok I guess i'm not clear on what csc is?

  10. bahrom7893
    • 5 years ago
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    Csc is Cosecant = 1/Sin

  11. anonymous
    • 5 years ago
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    oh ok this is making more sense now! Thank you!

  12. anonymous
    • 5 years ago
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    ok sorry one more time what is sin equal to then?

  13. bahrom7893
    • 5 years ago
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    Sin and Cos are basic units, everything can be broken down into them

  14. bahrom7893
    • 5 years ago
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    u can't break those down

  15. anonymous
    • 5 years ago
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    ok so say (1/1+sinx)+(1/1+cscx)=1 (1/1+sinx)+(1/1+(1/sin))=1 ok where do I go from there

  16. bahrom7893
    • 5 years ago
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    in the first parenthesis is it 1/(1+Sinx), i mean is 1+sinx in the bottom?

  17. anonymous
    • 5 years ago
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    yes sorry about that

  18. bahrom7893
    • 5 years ago
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    and btw is the 1+Cscx in the denominator too? always put parenthesis

  19. anonymous
    • 5 years ago
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    Yes sorry about that

  20. bahrom7893
    • 5 years ago
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    k workin on it now

  21. anonymous
    • 5 years ago
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    Thanks

  22. bahrom7893
    • 5 years ago
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    1/(1+Sinx) + 1/(1+Cscx) = 1/(1+Sinx) + 1/(1 + [1/Sinx])

  23. bahrom7893
    • 5 years ago
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    1/(1+Sinx) + {1 / [(Sinx+1)/Sinx]} = 1/(1+Sinx) + Sinx/(1+Sinx)

  24. bahrom7893
    • 5 years ago
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    Now common denominator as you can see is 1+Sinx => You can rewrite as: (1+Sinx)/(1+Sinx) = 1

  25. anonymous
    • 5 years ago
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    which narrows down to 1=1 right

  26. bahrom7893
    • 5 years ago
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    yeah

  27. anonymous
    • 5 years ago
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    awesome thanks

  28. anonymous
    • 5 years ago
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    ok what if I have ((1-2sin^2(θ))/(sin(θ)cos(θ)))=cot(θ)-tan(θ) I have (1-2sin^2θ)/(sinθ(cosθ/sinθ)=(cotθ-(sinθ/cosθ)? where do I go from there?

  29. bahrom7893
    • 5 years ago
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    WOW lol

  30. anonymous
    • 5 years ago
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    is that wrong?

  31. bahrom7893
    • 5 years ago
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    no its long lol

  32. anonymous
    • 5 years ago
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    oh well the bottom line is what I plugged in....

  33. bahrom7893
    • 5 years ago
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    hey can u write it out on a piece of paper and take a picture of it?

  34. anonymous
    • 5 years ago
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    Uhm ya I do online school so it's online on a word document ....

  35. bahrom7893
    • 5 years ago
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    okay email that to me

  36. bahrom7893
    • 5 years ago
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    bahrom.cfo@gmail.com

  37. anonymous
    • 5 years ago
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    ok...

  38. anonymous
    • 5 years ago
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    ok just sent it

  39. bahrom7893
    • 5 years ago
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    k lookin at it

  40. anonymous
    • 5 years ago
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    ?

  41. bahrom7893
    • 5 years ago
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    workin on it

  42. bahrom7893
    • 5 years ago
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    [1 - 2 Sin^2(t)] / [Sint*Cost] = Cost/Sint - Sint/Cost

  43. bahrom7893
    • 5 years ago
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    [1 - 2 Sin^2(t)] / [Sint*Cost] = [Cos^2(t) - Sin^2(t)]/[Sint*Cost]

  44. bahrom7893
    • 5 years ago
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    subtract [Cos^2(t) - Sin^2(t)]/[Sint*Cost] from both sides

  45. bahrom7893
    • 5 years ago
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    [1-2Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] = 0

  46. anonymous
    • 5 years ago
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    Ok I'm lost...

  47. bahrom7893
    • 5 years ago
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    where?

  48. anonymous
    • 5 years ago
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    Sorry I'm probably frustrating you.... how did you get it equal to 0?

  49. bahrom7893
    • 5 years ago
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    I subtracted [Cos^2(t) - Sin^2(t)]/[Sint*Cost] from both sides

  50. bahrom7893
    • 5 years ago
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    [Cos^2(t) - Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] = 0 on the left

  51. bahrom7893
    • 5 years ago
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    get it?

  52. bahrom7893
    • 5 years ago
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    I meant on the right side: [Cos^2(t) - Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] is 0

  53. anonymous
    • 5 years ago
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    Oh oh ok.. Ihad the right side... I get it thank you I'm messing with the number 10 problem now

  54. bahrom7893
    • 5 years ago
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    k so

  55. bahrom7893
    • 5 years ago
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    [1-2Sin^2(t)]/[Sint*Cost] - [Cos^2(t) - Sin^2(t)]/[Sint*Cost] = 0 The common denominator is Sint Cost: [1 - 2*Sin^2(t) - Cos^2(t) + Sin^2(t)] / [Sin(t)*Cos(t)] = 0

  56. anonymous
    • 5 years ago
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    ok

  57. bahrom7893
    • 5 years ago
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    Simplify the top: ( 1 - Sin^2(t) - Cos^2(t)) / (Sint*Cost) = 0

  58. bahrom7893
    • 5 years ago
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    notice that the top is also: ( 1 - [Sin^2(t) + Cos^2(t)]) / (Sint*Cost) = 0

  59. bahrom7893
    • 5 years ago
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    (1-1)/(Sint*Cost) = 0 0/(Sint*Cost) = 0 0 = 0 VOILA!

  60. bahrom7893
    • 5 years ago
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    and can u become my fan lol, im tryin to get more fans

  61. bahrom7893
    • 5 years ago
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    For the last one, square both sides

  62. anonymous
    • 5 years ago
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    Ha ha I already did! Thank you so much if I could be your fan more than once I would thank you so much!

  63. bahrom7893
    • 5 years ago
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    LOl its fine

  64. bahrom7893
    • 5 years ago
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    (1+Sint)/(1-Sint) = (1+Sint)^2 / Cos^2(t), u can ignore the abs value cuz Cos^2 is always positive

  65. anonymous
    • 5 years ago
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    ahh wait ok just checking the stuff youjust showed me.....that was #4 right?

  66. bahrom7893
    • 5 years ago
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    yeah

  67. anonymous
    • 5 years ago
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    ok I had it down there but wasn't sure... ok I'm following ya now

  68. bahrom7893
    • 5 years ago
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    okay can i work on ur last one in a couple of mins? i promised to help someone

  69. anonymous
    • 5 years ago
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    Ya no problem I'll mess around with it and see how far I get thanks

  70. bahrom7893
    • 5 years ago
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    okay how far r u?

  71. anonymous
    • 5 years ago
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    not much farther... I started where we left off but got way lost... I think I started dividing things that shouldnt be divided

  72. anonymous
    • 5 years ago
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    This is what I have.... wait did you say square both sides?

  73. bahrom7893
    • 5 years ago
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    yeah

  74. bahrom7893
    • 5 years ago
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    you'll get: (1+Sint)/(1-Sint) = (1+Sint)^2/Cos^2(t)

  75. anonymous
    • 5 years ago
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    ok I have that

  76. bahrom7893
    • 5 years ago
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    Okay now move to th(1+Sint)/(1-Sint) - (1+Sint)^2/Cos^2(t) = 0

  77. bahrom7893
    • 5 years ago
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    Divide everything by (1+Sint)

  78. anonymous
    • 5 years ago
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    ok so then... (1-sint)-cos^2(t)?

  79. bahrom7893
    • 5 years ago
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    no

  80. bahrom7893
    • 5 years ago
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    1/(1-Sint) = (1+Sint)/Cos^2(t)

  81. anonymous
    • 5 years ago
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    ugh... alright oh ok

  82. anonymous
    • 5 years ago
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    sorry I didn't mean to take up this much of your time

  83. bahrom7893
    • 5 years ago
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    1 - Sin(t) = (Cos^2(t))/(1+Sin(t))

  84. bahrom7893
    • 5 years ago
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    Multiply both sides by (1+Sin(t)): (1-Sint) (1+Sint) = Cos^2(t)

  85. bahrom7893
    • 5 years ago
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    Recognize the left side as: 1 - Sin^2(t) = Cos^2(t) 1 = Sin^2(t) + Cos^2(t) 1 = 1 <---FINALLY DONE WITH UR HW!!!

  86. anonymous
    • 5 years ago
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    I'm really sorry... thank you and good luck with your fan contest! thank you so very much!

  87. bahrom7893
    • 5 years ago
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    lol its not a contest, its just for fun lol and ur welcome!

  88. anonymous
    • 5 years ago
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    I wish I understood it more but that's the faultyness of online school... I get the jist! thanks

  89. bahrom7893
    • 5 years ago
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    no im off to bed nite. i woke up at 6 am today, left at 6:30 for college got home around 5:30pm, im really tired

  90. anonymous
    • 5 years ago
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    I hear ya! Night thanks again!

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