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Complex Sin value greater than 1

  1. Apr 12, 2014 #1
    My Complex Euler stuff is a bit rusty but I'm working on a problem (that I won't mention here) but I'm after an explanation as to how:

    Sin (theta) = 3.08 and so
    Cos (theta) = -i*2.91

    What's throwing me is that I know Sin can only have a max at 1 trigonometrically (obviously) so I'm assuming 3 is something to do with like 3 rotations, but I could be wrong. I was also thinking like the 0.08 would be 4.6 degrees for theta. And I tried rearranging e^i*theata = CiS as
    e^i*theta - sin (theta) = -i*cos(theta) but that doesn't help.

    Any explanations?

    Thanks a lot!!
  2. jcsd
  3. Apr 12, 2014 #2
    I'm not sure what you're asking. There is no theta such that sin(theta) = 3.08 (to my knowledge) and it is completely wrong to say that -i*2.91 is "less than" 1.
  4. Apr 12, 2014 #3


    The cosine function will never give you a complex number.
  5. Apr 12, 2014 #4
    He is talking about complex arguments. I think we can all agree that cos(5i + 2) is complex.
  6. Apr 12, 2014 #5
    Right yes, forgot about the general a+bi case
  7. Apr 12, 2014 #6
    We can say that:
    For real a:
    sin(a) is in [-1,1]

    For imaginary x:
    sin(bi) is imaginary, and so there is no well defined notion of a less than or greater than comparison to real numbers.

    For complex a + bi (that are not one of the above cases)
    sin(a + bi) is complex, and so there is no well defined notion of a less than or greater than comparison to real numbers.

    In any case, sin(x) is never greater than 1.
  8. Apr 12, 2014 #7


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    ^not quite
    sin(a+b i)
    is zero when a,b=0
    real when b or cos(a)=0
    imaginary when sin(a)=0

    and complex otherwise

    |e^(i theta)|=1
    only holds for theta real
    if theta is complex it does not hold
    still holds but complex squares can be negative

    Sin (theta) = 3.08 and so
    Cos (theta) = -i*2.91
    where n is any (rational) integer
    this ignores round off 3.08^2-2.91^2=1.0183
    when it should be 1
  9. Apr 12, 2014 #8
    Oh, OK. Sorry for the misinformation then.
  10. Apr 13, 2014 #9
    Thanks for the great replies everyone, I'm going to have another read through them again but the general sense I'm getting is that the theata can't be broken up so that the amplitude of sin is less than 1. This leads me to think that I'm going to have to treat Sin(theta) and Cos(theta) as variables completely.

    lurflurf, I should have specified that I'm not trying to find theta necessarily, more to deduce how we got to the Cos value. (But out of pure interest, what relation did you use to show theta=(2n+1)pi/2+i*arccosh(3.08)?)
    I tried the trig relation you reminded me of and rearranged for Cos but it didn't make much sense:
    sin^2 (theta) + cos^2 (theta) = 1 then
    cos(theta) = SQRT(1 - sin^2(theta)) = i*2.913 Which is correct but positive though.
    Surprisingly I'm just not seeing how:

    Can get me from Sin (theta) = 3.08 to Cos (theta) = -i*2.91....

    Is there any significance of it being greater than 1, like the rotations or something like I thought, or just some mathematical freakery?
  11. Apr 13, 2014 #10


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    sin^2 (theta) + cos^2 (theta) = 1
    holds when theta is complex
    the thing about sin(theta)<=1 follows from that and the assumption that sin(theta) and cos(theta) are real
    Sin (theta) = 3.08 tells us
    cos^2 (theta)=1-sin^2 (theta)=-8.8464
    you said Cos (theta) = -i*2.91
    Cos (theta) = i*2.91
    is also possible when sin(theta)=3.08
    if Cos (theta) = i*2.91
    where n is any (rational) integer
    knowing sin(theta) or cos(theta) allows us to find the other up to sign
  12. Apr 13, 2014 #11
    Oh I see.
    But so you're saying Cos (theta) = -i*2.91 was wrong, in that it should have been positive in the first place?

  13. Apr 13, 2014 #12


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    cos (theta) = -i*2.91
    cos (theta) = i*2.91
    are both possible given
    we cannot know which value we have without more information
  14. Apr 13, 2014 #13
    Ooh, right. Cheers
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