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anonymous

  • one year ago

cot2x + csc2x = 2csc2x - 1 my answer > cot^2+csc^4*x^3

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  1. anonymous
    • one year ago
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    Teacher:You cannot take terms across the = when doing a proof. You must show that the LHS=RHS. See if you can use cot^2(x)+1 = csc^2(x) to help you do this.??

  2. anonymous
    • one year ago
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    @freckles

  3. freckles
    • one year ago
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    so those things in your problem are not double angles right?

  4. anonymous
    • one year ago
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    idont thinks so no there not

  5. xapproachesinfinity
    • one year ago
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    looks like square not double angle that way the left hand side is correct

  6. anonymous
    • one year ago
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    yes

  7. freckles
    • one year ago
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    so the problem is: \[\cot^2(x)+\csc^2(x)=2\csc^2(x)-1\]

  8. anonymous
    • one year ago
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    yes

  9. xapproachesinfinity
    • one year ago
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    this problem wants you to use the id 1+cot^2=csc^2

  10. freckles
    • one year ago
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    well let's play with this: \[\cot^2(x)+1=\csc^2(x)\] subtract 1 on both sides

  11. freckles
    • one year ago
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    this will isolate the cot^2(x) term

  12. freckles
    • one year ago
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    in then you will be able to write left hand side in terms of csc

  13. anonymous
    • one year ago
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    umm okkk

  14. freckles
    • one year ago
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    do this one step: subtract 1 on both sides of cot^2(x)+1=csc^2(x)

  15. anonymous
    • one year ago
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    cot^2(x)csc^2 right

  16. freckles
    • one year ago
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    what happen to the equation

  17. freckles
    • one year ago
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    if I asked you to subtract 1 on both sides of x+1=y what would you get after just doing that one step

  18. anonymous
    • one year ago
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    you would get x=y

  19. freckles
    • one year ago
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    i'm asking you to subtract one on both sides not one side remember whatever you do to one side of the equation you have to do to the other

  20. freckles
    • one year ago
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    x+1=y subtract 1 on both sides (x+1)-1=y-1 x+1-1=y-1 x=y-1

  21. anonymous
    • one year ago
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    ohh ok

  22. freckles
    • one year ago
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    \[\cot^2(x)+1=\csc^2(x) \\ \text{ subtract 1 on both sides here }\]

  23. anonymous
    • one year ago
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    cot^2(X)=CSC^2(X)-1

  24. freckles
    • one year ago
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    right \[\cot^2(x)+\csc^2(x) \text{ was your left hand side } \\ \text{ replace} \cot^2(x) \text{ with } \csc^2(x)-1\]

  25. freckles
    • one year ago
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    \[(\csc^2(x)-1)+\csc^2(x) \\ \text{ you can drop ( ) } \\ \csc^2(x)-1+\csc^2(x)\] do you think you can finish here:

  26. freckles
    • one year ago
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    you know how to combine like terms right?

  27. anonymous
    • one year ago
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    no

  28. freckles
    • one year ago
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    examples of combining like terms: 5+5=2(5)=10 a+a=2a x^2+x^2=2x^2 log(x)+log(x)=2log(x) sin(x)+sin(x)=2sin(x) sin^2(x)+sin^2(x)=2sin^2(x) sin^2(x)+4sin^2(x)=5sin^2(x) can you find this: csc^2(x)+csc^2(x)=?

  29. freckles
    • one year ago
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    or not find but simplify

  30. anonymous
    • one year ago
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    (2*x)*csc^2

  31. freckles
    • one year ago
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    what happen to csc^2 's angle part?

  32. anonymous
    • one year ago
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    ??

  33. freckles
    • one year ago
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    csc^2 has no meaning csc^2(x) has meaning

  34. freckles
    • one year ago
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    something has to be plugged into the csc^2 function to actually have meaning csc^2 alone means nothing

  35. anonymous
    • one year ago
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    so (2*x)*csc^2(X)

  36. freckles
    • one year ago
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    so I think you meant to write 2csc^2(x) this does not mean csc^2 times x it means csc^2 of x like means x is being plugged into the csc^2( ) function

  37. freckles
    • one year ago
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    where does the extra x come from

  38. freckles
    • one year ago
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    \[\csc^2(x)+\csc^2(x)=2 \csc^2(x) \]

  39. anonymous
    • one year ago
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    ohh i see what you mena venr mind that last statment lol

  40. freckles
    • one year ago
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    anyways do you have any questions on this one before I take off for a bit

  41. anonymous
    • one year ago
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    id thnx alot

  42. freckles
    • one year ago
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    np

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