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anonymous

  • one year ago

sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x

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  1. anonymous
    • one year ago
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    Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

  2. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } +\frac{ \sin(x) }{ 1 -cosine(x) }\] what is the common denominator ?

  3. anonymous
    • one year ago
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    1-cosine(x)?

  4. anonymous
    • one year ago
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    but the second one is sinx/1+cosx

  5. Nnesha
    • one year ago
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    well the common denominator is 1-cosine(x) obviously both denominator are same u will get same numerator (sinx)\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } \]

  6. Nnesha
    • one year ago
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    huh uh-oh are you sure check your question again

  7. anonymous
    • one year ago
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    yea i checked it 1+cosx

  8. Nnesha
    • one year ago
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    http://prntscr.com/7vfzly

  9. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } +\frac{ \sin(x) }{ 1 +cosine(x) }\] so common denominator is (1-cosinex)(1+cosinex)

  10. anonymous
    • one year ago
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    but the picture of the equation in the book says 1+cosx

  11. Nnesha
    • one year ago
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    alright you typed it wrong

  12. anonymous
    • one year ago
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    it did that itself it copy pasted that way lol

  13. anonymous
    • one year ago
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  14. Nnesha
    • one year ago
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    really ? :o you didin't type it ?

  15. Nnesha
    • one year ago
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    shocking how that can translate :o :/

  16. anonymous
    • one year ago
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    lol no that was weird tho right?

  17. Nnesha
    • one year ago
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    \[\huge\rm \frac{ sinx(1+cosx) +sinx(1-cosx)}{ (1-cosx)(1+cosx)}\] common denominator is (1-cosx)(1+cosx) and multiply numerator of first fraction by denominator of 2nd fraction *and multiply numerator of 2nd fraction by denominator of first fraction

  18. Nnesha
    • one year ago
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    ye ....

  19. anonymous
    • one year ago
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    and then you cancel out?

  20. Nnesha
    • one year ago
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    no no NO! no then u will get the original question

  21. Nnesha
    • one year ago
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    distribute!

  22. anonymous
    • one year ago
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    ohhh

  23. anonymous
    • one year ago
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    wait what

  24. Nnesha
    • one year ago
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    yes familiar withe the foil method ? apply that at the denominator

  25. anonymous
    • one year ago
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    oh okay so i leave the numerator alone righ tnow

  26. anonymous
    • one year ago
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    1-cos2x?

  27. Nnesha
    • one year ago
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    yep right now distribute both parentheses by sinx 't the numerator

  28. anonymous
    • one year ago
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    what? that last sentence you said threw me off i guess the wording did

  29. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \color{reD}{sinx}(1+cosx) +\color{red}{sinx}(1-cosx)}{ (1-cosx)^2}\] distribute both parentheses by red sin x

  30. anonymous
    • one year ago
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    1sinx+sinxcosx+1sinx-sinxcosx

  31. anonymous
    • one year ago
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    2sinx/1-cos2x

  32. Nnesha
    • one year ago
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    and cos^2x equal to what ? do u remember sin/cos identities ?

  33. anonymous
    • one year ago
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    csc?

  34. Nnesha
    • one year ago
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    nope

  35. Nnesha
    • one year ago
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    special identity \[\huge\rm sin^2x + \cos^2 x = 1\] solve this for cos^2x

  36. anonymous
    • one year ago
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    some codes came out for when you said special identity nothing else?

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

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

  39. Nnesha
    • one year ago
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    yes right so replace cos^2x by 1-sin^2x

  40. anonymous
    • one year ago
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    okay

  41. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \color{reD}{2sinx}}{ 1-(1-sin^2x)}\] distribute parentheses by negative one

  42. anonymous
    • one year ago
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    its coding

  43. Nnesha
    • one year ago
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    ahh 2sinx over 1-(1-sin^2x)

  44. anonymous
    • one year ago
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    sorry lol im being a pain

  45. anonymous
    • one year ago
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    2sinx/sin^2x

  46. Nnesha
    • one year ago
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    yes sin^2x is same as sinx times sin x so 2sin over sinx sin x simplfy

  47. anonymous
    • one year ago
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    1/sin2x

  48. Nnesha
    • one year ago
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    is it sin^2 x ?

  49. anonymous
    • one year ago
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    yea but with one over it

  50. Nnesha
    • one year ago
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    no what about 2 ? it's 2sinx over sinx times sin x

  51. Nnesha
    • one year ago
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    |dw:1437522158545:dw|

  52. anonymous
    • one year ago
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    it all cancels out to 1/sinx

  53. Nnesha
    • one year ago
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    no what about 2 ??|dw:1437522218631:dw|

  54. anonymous
    • one year ago
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    im so lost i dont like this number two lol

  55. Nnesha
    • one year ago
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    just look at that drawing who kidnapped that 2 ?

  56. anonymous
    • one year ago
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    the sinx? i have no idea

  57. Nnesha
    • one year ago
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    sin x cancels out right ?? so where is 2 ? look at there

  58. anonymous
    • one year ago
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    2 over sinx

  59. Nnesha
    • one year ago
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    ahh there ugo!

  60. Nnesha
    • one year ago
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    i was about ...to .. put u in jail how....darrrrrrrrrrrrrre u kidnapped 2 :/ ;) hey gtg gO_OD luck!

  61. anonymous
    • one year ago
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    haha okay but what do i do to finish?

  62. Nnesha
    • one year ago
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    well 2 over sinx is same as 2 times 1/sinx and 1/sinx = csc 2 times csc

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