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anonymous

  • one year ago

Verify the identity. cotangent of x to the second power divided by quantity cosecant of x plus one equals quantity one minus sine of x divided by sine of x

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  1. triciaal
    • one year ago
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    |dw:1443371271009:dw|

  2. triciaal
    • one year ago
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    the original as written is not clear.

  3. anonymous
    • one year ago
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    1-2sin^2 x/ sin^2 x / 1+sin x/ sin x = 1- sin^2 x/ sinx * 1+ sinx

  4. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \cot^2 (x) }{ \csc(x) +1 } =\frac{ 1-\sin(x)}{ \sin(x)}\] like this ?

  5. anonymous
    • one year ago
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    I know if I factor 1-(sinx)^2 I'll get 1+sinx and 1-sinx

  6. anonymous
    • one year ago
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    yes like that^

  7. Nnesha
    • one year ago
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    alright gimme a sec

  8. Nnesha
    • one year ago
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    are you working on left or right side ?

  9. anonymous
    • one year ago
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    left

  10. Nnesha
    • one year ago
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    alright first we can rewrite cot and csc in terms of sin and cos

  11. Nnesha
    • one year ago
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    csc = 1/sin right so cot = ??

  12. anonymous
    • one year ago
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    1/tan?

  13. anonymous
    • one year ago
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    cos/sin

  14. Nnesha
    • one year ago
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    yeah but we need to write in terms of sin cos

  15. Nnesha
    • one year ago
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    yes right \[\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1 }{ \sin(x) }+1} }\] now first deal with the denominator (red part)

  16. Nnesha
    • one year ago
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    \[\frac{ 1 }{ \sin(x) }+1\] find the common denominator

  17. anonymous
    • one year ago
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    well 1/ sin = csc and 1+ csc = cot^2?

  18. anonymous
    • one year ago
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    I'm confused.

  19. Nnesha
    • one year ago
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    yes right you can use the identity but i found the other one easy

  20. Nnesha
    • one year ago
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    but if you use the identity you still have to write csc in terms of sin and cos

  21. anonymous
    • one year ago
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    so what do I do, I know csc is 1/sin

  22. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1 }{ \sin(x) }+1} }\] now first deal with the denominator (red part) we need to find common denominator 1 is same as 1/1 so \[\frac{ 1}{ \sin(x)} +\frac{ 1 }{ 1 }\] common denomiantor is sin x right

  23. Nnesha
    • one year ago
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    \[\large\rm \frac{ 1 }{ \color{blue}{\sin(x)} }+\frac{ 1 }{ \color{red}{1}}\] \[\huge\rm \frac{ \color{red}{1}(1) +\color{blue}{sinx }(1)}{ \sin(x) }\] when we find common denominator we should multiply first fraction with the denominator of 2nd one and multiply the numerator of 2nd fraction with the denominator of first fraction

  24. Nnesha
    • one year ago
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    let me know if you have a question about that part ??^^^

  25. anonymous
    • one year ago
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    No it makes sense.

  26. Nnesha
    • one year ago
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    alright \[\huge\rm \frac{ \frac{ \cos^2(x) }{ \sin^2(x) } }{ \color{reD}{\frac{ 1+sin(x) }{ \sin(x) }} }\] now change division to multiplication multiply first fraction with the `reciprocal` of the 2nd fraction

  27. Nnesha
    • one year ago
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    here is an example \[\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } } =\frac{ a }{ b} \times \frac{ d }{ c }\]

  28. Nnesha
    • one year ago
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    no multiply top fraction with the reciprocal of bottom one what's reciprocal of 1+sinx/sin ???

  29. anonymous
    • one year ago
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    sin/1+sinx

  30. Nnesha
    • one year ago
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    right so multiply \[\huge\rm \frac{ \cos^2(x) }{\sin^2(x) } \times \frac{ \sin(x) }{1+\sin(x)}\]

  31. Nnesha
    • one year ago
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    sin^2 is same as sin times sin so you cancel it out

  32. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \cos^2(x) }{\color{reD}{sin^2(x)} } \times \frac{ \sin(x) }{1+\sin(x)}\] sin^2 is same as sin times sin so you cancel it out \[\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times sinx} } \times \frac{ \sin(x) }{1+\sin(x)}\]

  33. anonymous
    • one year ago
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    so does that mean sinx cancels out?

  34. Nnesha
    • one year ago
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    yes right

  35. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times \cancel{sinx}} } \times \frac{\cancel{ \sin(x)} }{1+\sin(x)}\]

  36. Nnesha
    • one year ago
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    now use the identity cos^2(x) = ???

  37. anonymous
    • one year ago
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    cos^2 = cosx * cosx?

  38. Nnesha
    • one year ago
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    yes but we need identity

  39. Nnesha
    • one year ago
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    sin^2(x) +cos^2(x) =1 solve for cos^2(x)

  40. anonymous
    • one year ago
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    oh wait its a path. identity.

  41. Nnesha
    • one year ago
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    that's the identity

  42. Nnesha
    • one year ago
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    ye

  43. anonymous
    • one year ago
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    That's it?

  44. Nnesha
    • one year ago
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    well ye just replace cos^2 with the identity then done

  45. Nnesha
    • one year ago
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    cos^2x = ???

  46. anonymous
    • one year ago
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    sin^2 + cos^2

  47. anonymous
    • one year ago
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    cos^2=sin^2?

  48. Nnesha
    • one year ago
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    how would you cancel out sin^2 from left side \[\huge\rm sin^2x+\cos^2=1\] ?

  49. Nnesha
    • one year ago
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    subtract/add/divide /mutliply both sides by sin^2 x ?

  50. anonymous
    • one year ago
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    I don't understand, how? ^

  51. Nnesha
    • one year ago
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    yes i'm asking how would you cancel out sin^2x from left its just like simple equation

  52. Nnesha
    • one year ago
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    lets say we have x +y = 1 how would you solve for y ?

  53. anonymous
    • one year ago
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    you move x to the other side right?

  54. Nnesha
    • one year ago
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    yes right what would you get ?

  55. anonymous
    • one year ago
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    not enough info to solve, because x=?

  56. Nnesha
    • one year ago
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    well lets say 2+x =1 how would you solve for x ?

  57. anonymous
    • one year ago
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    -2 both sides x = -1

  58. Nnesha
    • one year ago
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    yes right subtract 2 both sides so for this equation subtract sin2x both sides \[\huge\rm sin^2x+\cos^2=1\]

  59. Nnesha
    • one year ago
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    |dw:1443375444006:dw| what would you get at right side ?

  60. anonymous
    • one year ago
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    cos^2= 1-sin^2

  61. Nnesha
    • one year ago
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    yes right now u can replace cos^2x with that \[\huge\rm \frac{ \cos^2(x) }{\color{reD}{sinx \times \cancel{sinx}} } \times \frac{\cancel{ \sin(x)} }{1+\sin(x)}\] \[\huge\rm \frac{ 1-sin^2(x)}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}\]

  62. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @clairvoyant1 I know if I factor 1-(sinx)^2 I'll get 1+sinx and 1-sinx \(\color{blue}{\text{End of Quote}}\) now as u said you can write 1-sin^2x as (1+sinx)(1-sinx)

  63. Nnesha
    • one year ago
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    \[\huge\rm \frac{ (1+sin(x))(1-sin(x))}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}\]

  64. Nnesha
    • one year ago
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    you can cancel out something right after that done!

  65. Nnesha
    • one year ago
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    got it ?? :=)

  66. anonymous
    • one year ago
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    Computer froze logged in on my kindle Lol. Not quite

  67. anonymous
    • one year ago
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    @Nnesha

  68. Nnesha
    • one year ago
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    \[\huge\rm \frac{ (1+sin(x))(1-sin(x))}{\color{reD}{sinx } } \times \frac{{ 1}}{1+\sin(x)}\] what can you cancel out ?

  69. anonymous
    • one year ago
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    1+sin(x)

  70. Nnesha
    • one year ago
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    \[\large\rm \frac{ (1-sin(x))\cancel{(1+sin(x))} }{\color{reD}{sinx } } \times \frac{{ 1}}{\cancel{1+\sin(x)}}\]

  71. Nnesha
    • one year ago
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    that's it

  72. Nnesha
    • one year ago
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    both sides are the same right ?

  73. anonymous
    • one year ago
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    Ok could you do a quick recap of what we just went over please.

  74. anonymous
    • one year ago
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    @Nnesha

  75. Nnesha
    • one year ago
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    first rewrite cot and csc in terms of sin and cos then we found the common denominator change division to multiplication use the identity cos^2=1-sin^2 that's it

  76. Nnesha
    • one year ago
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    if you need the work you can reread the post :=) feel free to ask q

  77. anonymous
    • one year ago
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    Thank You so much for your patience and help :)

  78. Nnesha
    • one year ago
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    np :=)

  79. anonymous
    • one year ago
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    Verify the identity. cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x

  80. anonymous
    • one year ago
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    @Nnesha

  81. anonymous
    • one year ago
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    Can you help me with this one.

  82. Nnesha
    • one year ago
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    i'm not sure about that one sorry i just know the basic :(

  83. Nnesha
    • one year ago
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    make new post so other people will b able to help chu

  84. anonymous
    • one year ago
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    Find the domain of the given function. f(x) = square root of quantity x plus three divided by quantity x plus eight times quantity x minus two. a) x > 0 b) All real numbers c) x ≥ -3, x ≠ 2 d) x ≠ -8, x ≠ -3, x ≠ 2

  85. Nnesha
    • one year ago
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    can you post the pic ?

  86. anonymous
    • one year ago
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    Verify the identity. cos 4x + cos 2x = 2 - 2 sin2 2x - 2 sin2 x @campbell_st

  87. campbell_st
    • one year ago
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    well cos(4x )= 1 - 2sin^2(2x) and cos(2x) = 1- 2sin^2(x) now make the substitution and simplify

  88. anonymous
    • one year ago
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    @campbell_st how would I do that?

  89. campbell_st
    • one year ago
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    exactly what it says... replace cos(4x) with the right hand side of the expression then the same with cos(2x) and collect like terms. what would probably be best, is to post it as a new question... as there is an lot of information that precedes the question you asked.

  90. anonymous
    • one year ago
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    alright I'll post it as a new question

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