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SugarRainbow

  • 2 years ago

simplify tan x cos x

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  1. qpHalcy0n
    • 2 years ago
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    Remember tangent is just sin(x)/cos(x)

  2. SugarRainbow
    • 2 years ago
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    so its cos x (sin x / cos x) ?

  3. qpHalcy0n
    • 2 years ago
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    Correct, now what happens when you go ahead and multiply the cosine?

  4. SugarRainbow
    • 2 years ago
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    sin x

  5. qpHalcy0n
    • 2 years ago
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    That's it. The cosines cancel.

  6. SugarRainbow
    • 2 years ago
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    okay cool what bout sec y sin (pi/2-y)? IDK how to do this at all

  7. qpHalcy0n
    • 2 years ago
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    \[\sec(y)\sin(\frac{\pi}{2-y})\] ??

  8. SugarRainbow
    • 2 years ago
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    but isnt it like pi/2 - y not pi/ 2-y

  9. hartnn
    • 2 years ago
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    sin(pi/2-y)=cos y sec y= 1/cos y

  10. qpHalcy0n
    • 2 years ago
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    Ok, so rewrite secant as 1/cos

  11. qpHalcy0n
    • 2 years ago
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    Then you can use a fancy pants trig identity.

  12. SugarRainbow
    • 2 years ago
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    cos y 1/cos y

  13. qpHalcy0n
    • 2 years ago
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    |dw:1347861835303:dw| Right?

  14. SugarRainbow
    • 2 years ago
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    i think so i was kinda looking at what hartnns post too so couldn't you do it like|dw:1347861948541:dw|

  15. hartnn
    • 2 years ago
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    sec y =1/ cos y is in numerator, so it would be :|dw:1347862109049:dw|

  16. qpHalcy0n
    • 2 years ago
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    No, well what you would have is a multiplication, the cosines would cancel.

  17. SugarRainbow
    • 2 years ago
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    oh hey you know how tan x= sinx/ cos x well what are the rest of those cuz it might help me understand better like how we getting cos y when its sin y

  18. hartnn
    • 2 years ago
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    ok, so u want all trignometric formulas ? for converting one ratio to other ?

  19. SugarRainbow
    • 2 years ago
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    yes please

  20. hartnn
    • 2 years ago
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    i will list some : \(\huge sin x =1/cosec x \implies cosec x=1/sin x \\\huge cos x = 1/sec x \implies secx=1/cos x \\\huge tan x=1/cot x \implies cot x =1/tanx \\\huge tan x=sin x/cos x \\\huge cot x=cos x/sinx \\ \huge sin^2x+cos^2x=1 \\\huge sec^2x=1+tan^2x\\\huge cosec^2x=1+cot^2x\)

  21. hartnn
    • 2 years ago
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    if u want sin x when cos x is given or if u want cos x when sin x is given, u use: \(sin^2x+cos^2x=1\)

  22. SugarRainbow
    • 2 years ago
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    okay

  23. SugarRainbow
    • 2 years ago
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    so if its 1+tan^2x/csc^2x that would simplify to sec^2x/csc^2

  24. hartnn
    • 2 years ago
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    yes. or is it ? \(\large1+\frac{tan^2x}{cosec^2x}\quad?\)

  25. hartnn
    • 2 years ago
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    if it is \(\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)

  26. SugarRainbow
    • 2 years ago
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    but you put that cos^2x=1+cot^2x

  27. hartnn
    • 2 years ago
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    no the last one is cosec^2 x =1+cot^2 x not cos^2x=1+cot^2x

  28. hartnn
    • 2 years ago
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    csc=cosec

  29. SugarRainbow
    • 2 years ago
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    oh yeah huh cosec is csc not cos i forgot for a minute lol

  30. SugarRainbow
    • 2 years ago
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    so is it 1+ tan^2x/csc^2x or sec^2x/csc^2 and why

  31. hartnn
    • 2 years ago
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    what exactly is your question ? is this : (1+ tan^2x)/csc^2x

  32. hartnn
    • 2 years ago
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    (1+ tan^2x)=sec^2 x from the formula i gave u.

  33. SugarRainbow
    • 2 years ago
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    no i saying is it my answer or yours and why (idk how to put it on my computer like you did so that was the best i could do)

  34. hartnn
    • 2 years ago
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    to tell u what is your answer, i need to know your exact question, that is why i asked whether it is: (1+ tan^2x)/csc^2x

  35. SugarRainbow
    • 2 years ago
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    oh yeah it's that and i need to simplify it

  36. hartnn
    • 2 years ago
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    ok,for that i have shown u the steps, tell me which step, u didn't understand: \(\large\frac{1+tan^2x}{cosec^2x}=\frac{sec^2x}{cosec^2x}=\frac{sin^2x}{cos^2x}=tan^2x\)

  37. SugarRainbow
    • 2 years ago
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    oh i get it now earlier you put it wierd so i didn't get it

  38. SugarRainbow
    • 2 years ago
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    okay now this one is cot x tan x so it would be (cosx/sinx)(sinx/cosx) correct?

  39. hartnn
    • 2 years ago
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    yes, or u could use cot x = 1/ tanx and u get tan x / tan x =1

  40. SugarRainbow
    • 2 years ago
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    wait because i have two of each would thay cancel out?

  41. hartnn
    • 2 years ago
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    yes, they would. sin x cos x / sin x cos x =1

  42. SugarRainbow
    • 2 years ago
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    oh okay just checking

  43. hartnn
    • 2 years ago
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    |dw:1347864987907:dw|

  44. hartnn
    • 2 years ago
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    okk.

  45. SugarRainbow
    • 2 years ago
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    how bout cot u sin u would that simplify to (cosu/sinu)sin u= cos u?

  46. hartnn
    • 2 years ago
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    thats absolutely correct! u are improving :)

  47. SugarRainbow
    • 2 years ago
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    hooray improvement!

  48. hartnn
    • 2 years ago
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    what next ?

  49. SugarRainbow
    • 2 years ago
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    1-cos^2 beta/sin beta and that would simplify to ooh okay i got nothin for this one :(

  50. hartnn
    • 2 years ago
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    is the numerator (1-cos^2 beta ) ??

  51. SugarRainbow
    • 2 years ago
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    yes

  52. hartnn
    • 2 years ago
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    from sin^2 beta +cos^2 beta =1 u have 1-cos^2 beta = sin^2 beta so u have : \(\large \frac{sin^2 \beta}{sin \beta}=\frac{sin \beta.sin\beta}{sin \beta}=?\)

  53. SugarRainbow
    • 2 years ago
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    sin beta?

  54. hartnn
    • 2 years ago
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    yup, but did u get all steps ?

  55. SugarRainbow
    • 2 years ago
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    yeah so is that another function 1-cos^2 beta = sin^2 beta

  56. hartnn
    • 2 years ago
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    see the list i have given. it has sin^2 x + cos^2 x =1 isn't it ? from here i got 1-cos^2 x = sin^2 x clear ?

  57. SugarRainbow
    • 2 years ago
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    oh so you just went backwards?

  58. hartnn
    • 2 years ago
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    whenever u see sin^2 x or cos^2 x try to use that formula for simplification.

  59. SugarRainbow
    • 2 years ago
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    okay now i have cos x- cos^3 x

  60. SugarRainbow
    • 2 years ago
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    like would that be cosx- (cosx)(cosx)/cosx?

  61. hartnn
    • 2 years ago
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    nopes. take cos x common from both the terms, what do u get ?

  62. hartnn
    • 2 years ago
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    cosx- (cosx)(cosx)(cosx)

  63. SugarRainbow
    • 2 years ago
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    so you get 1

  64. hartnn
    • 2 years ago
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    no! after taking cos x common, u get cos x (1-(cos x)(cos x) ) =cos x (1- cos^2 x) and u already know what is (1- cos^2 x)

  65. SugarRainbow
    • 2 years ago
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    sin^2x

  66. hartnn
    • 2 years ago
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    yes! so your final answer will be cos x sin^2x but did u get all the steps ?

  67. SugarRainbow
    • 2 years ago
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    yeah because you have coxx-(cosx)(cosx)(cosx) =cosx-1(cosx)=1-cos^2x=sin^2x

  68. hartnn
    • 2 years ago
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    i will draw it :|dw:1347866550463:dw|

  69. SugarRainbow
    • 2 years ago
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    oh okay then now my last question is sin^2 u + tan^2 u +cos^2 u/sec u

  70. hartnn
    • 2 years ago
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    is this entire thing (sin^2 u + tan^2 u +cos^2 u) in the numerator ?

  71. SugarRainbow
    • 2 years ago
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    yes

  72. hartnn
    • 2 years ago
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    again using sin^2 x +cos^2 x =1, u have numerator as (1+tan^2 x) got this ? and (1+tan^2 x) = sec^2 x.....see formula list. so u have now sec^2x/sec x = ??

  73. hartnn
    • 2 years ago
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    replace all of my 'x' with 'u'

  74. SugarRainbow
    • 2 years ago
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    oh yeah so now you have sec x!

  75. hartnn
    • 2 years ago
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    yup :)

  76. hartnn
    • 2 years ago
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    sec u ;)

  77. SugarRainbow
    • 2 years ago
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    oh yeah lol thanks for your help i going to sleep now so goodnight^-^

  78. hartnn
    • 2 years ago
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    a very good night to u and have sweet dreams :)

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