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Sinh / sin

  1. May 16, 2016 #1
    1. The problem statement, all variables and given/known data
    So this is a really simple problem and i know I'm missing something really obvious but i just can't spot it.

    Screen Shot 2016-05-16 at 15.48.10.png

    2. Relevant equations


    3. The attempt at a solution
    so in the second part above I get :
    e-x - ex/2i. However I don't get the next bit as the 2i on the denominator is now factored into the top equation to give -ie-x + iex/2. I don't get how the 2 equations are equivalent and what they've done to get the i from the denominator to the numerator. If anyone could explain how this happened that'd be great!
     
  2. jcsd
  3. May 16, 2016 #2

    fresh_42

    Staff: Mentor

    What is i?
     
  4. May 16, 2016 #3
    Do you know what ##1/i## is?
     
  5. May 16, 2016 #4
    i-1?
     
  6. May 16, 2016 #5

    fresh_42

    Staff: Mentor

    First answer my question and then use it to answer struggles'. ##i^{-1}## is simply another form of ##\frac{1}{i}## which doesn't really help.
     
  7. May 16, 2016 #6

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Suppose you didn't know what ##i## is
    but it seems that (from your second equals sign) ##\frac {1}{i}= -i##... what must ##i## satisfy?
     
  8. May 16, 2016 #7

    fresh_42

    Staff: Mentor

    How is it defined? What stands ##i## for? There must have been some explanation for it.
    Remark: Sorry, for confusing your names, struggles and axmls!
     
  9. May 16, 2016 #8
    Though it isn't obvious that this is the way to find this, here's a hint:
    $$\frac{1}{i} = \frac{1}{i} \frac{i}{i}$$
     
  10. May 16, 2016 #9
    so i = √-1 so i2 = -1. so 1/i must equal 1/(-1)1/2 = (-1)-1/2? is that what you were getting at fresh-_42?
    So axmls 1/i = i/i^2 = -i! Thanks both of you!
     
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