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LaddiusMaximus

  • 5 years ago

(sinx/(1+cosx)) +((1+cosx)/sinx) totally lost

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  1. anonymous
    • 5 years ago
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    What do you have to do? prove it?

  2. LaddiusMaximus
    • 5 years ago
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    yeah=2cscx

  3. myininaya
    • 5 years ago
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    combine the fractions by finding a common denominator

  4. LaddiusMaximus
    • 5 years ago
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    yeah i got that, but what do I use, sinx? or 1+cosx?

  5. anonymous
    • 5 years ago
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    I'm working on it

  6. anonymous
    • 5 years ago
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    The Common Denominator will be sinx(1+cosx)

  7. myininaya
    • 5 years ago
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    what do you mean? how to combine these fractions: 2/3+5/7

  8. LaddiusMaximus
    • 5 years ago
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    so the common denominator is (sinx)1+cosx?

  9. myininaya
    • 5 years ago
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    sinx(1+cosx)

  10. anonymous
    • 5 years ago
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    yes that is the common denom

  11. LaddiusMaximus
    • 5 years ago
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    okay so now I have (sinx^2x(1+cosx)/sinx(1+cosx)) + (sinx(2+cos^2x)/sinx(1+cosx))?

  12. myininaya
    • 5 years ago
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    you will have (sinx)^2+(1+cosx)^2 in the numerator and sinx(1+cosx) is the denominator

  13. LaddiusMaximus
    • 5 years ago
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    sin^2x on both sides?

  14. anonymous
    • 5 years ago
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    I think you should have this so far: \[\sin^2x + (1+cosx)^2/sinx(1+cosx)\]

  15. myininaya
    • 5 years ago
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    ok look at your first fraction. what does the bottom lack so that it has the same denominator as the 2nd. sinx, right?

  16. myininaya
    • 5 years ago
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    multiply both the top and bottom by sinx on the first fraction

  17. LaddiusMaximus
    • 5 years ago
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    hmm i had [\sin^2x(1+cosx)+sinx(2+\cos^2x)/sinx(1+cosx)\]

  18. myininaya
    • 5 years ago
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    now look at the 2nd fraction what does the bottom lack so that it has the same denominator as the first fraction, (1+cosx), right?

  19. myininaya
    • 5 years ago
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    so multiply the top and bottom of the 2nd fraction by (1+cosx). Now the fractions have the same denominator, you can write it as one fraction now

  20. LaddiusMaximus
    • 5 years ago
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    I follow you so far.

  21. myininaya
    • 5 years ago
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    so we have [sinx*sinx+(1+cosx)(1+cosx)]/[sinx(1+cosx)]

  22. myininaya
    • 5 years ago
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    [(sinx)^2+1+2cosx+(cosx)^2]/[sinx(1+cosx)]

  23. anonymous
    • 5 years ago
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    I think I got the complete answer/proof

  24. myininaya
    • 5 years ago
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    what is (sinx)^2+(cosx)^2

  25. anonymous
    • 5 years ago
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    =1

  26. myininaya
    • 5 years ago
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    lol gj, but i was laddius

  27. anonymous
    • 5 years ago
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    oops sorry

  28. myininaya
    • 5 years ago
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    but i was asking*

  29. myininaya
    • 5 years ago
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    do you still follow laddius

  30. LaddiusMaximus
    • 5 years ago
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    so the numerator becomes sin^2x + (1+cosx)^2?

  31. anonymous
    • 5 years ago
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    yes

  32. myininaya
    • 5 years ago
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    now foil that (1+cosx)^2

  33. anonymous
    • 5 years ago
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    exactly lol

  34. LaddiusMaximus
    • 5 years ago
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    so now its 1+2cosx+cosx^2?

  35. anonymous
    • 5 years ago
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    yes!

  36. myininaya
    • 5 years ago
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    +(sinx)^2

  37. myininaya
    • 5 years ago
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    now do you see (sinx)^2+(cosx)^2 anywhere in that numerator

  38. LaddiusMaximus
    • 5 years ago
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    ahhhhh pythagorean

  39. myininaya
    • 5 years ago
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    what do you have now?

  40. LaddiusMaximus
    • 5 years ago
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    so now I have 2+2cosx/sinx(1+cosx)

  41. myininaya
    • 5 years ago
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    ok what does the numerator terms have in common

  42. LaddiusMaximus
    • 5 years ago
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    2?

  43. anonymous
    • 5 years ago
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    HAHA ignore my last 3 posts..

  44. myininaya
    • 5 years ago
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    does anything cancel?

  45. myininaya
    • 5 years ago
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    jp?

  46. LaddiusMaximus
    • 5 years ago
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    the 2s cancel, right?

  47. myininaya
    • 5 years ago
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    no the 1+cosx do

  48. myininaya
    • 5 years ago
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    so you have 2/sinx

  49. myininaya
    • 5 years ago
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    which is ....

  50. LaddiusMaximus
    • 5 years ago
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    I have 2+2cosx in the numerator

  51. LaddiusMaximus
    • 5 years ago
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    cosx+cosx= 2cosx, right?

  52. myininaya
    • 5 years ago
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    didnt you say the numerator terms had a 2 in common so factor that out in you have 2(1+cosx)/[sinx(1+cosx)]

  53. LaddiusMaximus
    • 5 years ago
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    crap then the (1+cosx) cancel out and im left with 2cscx

  54. myininaya
    • 5 years ago
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    BINGO!

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