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thrustfault Group Title

Let f(x) = sin(x) + icos(x). Show that |f(x)| = 1. I'm not even sure where to start here...

  • 2 years ago
  • 2 years ago

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  1. khurramshahzad Group Title
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    hey its too much easy... |f(x)|=sqrt((1)^2+(1)^2)=sqrt(2)

    • 2 years ago
  2. khurramshahzad Group Title
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    here we will take its real part 1 will cos(x) and also 1 from imagnary part i(sinx) hey u have written wrong equation.. its cos(x)-isin(x)

    • 2 years ago
  3. thrustfault Group Title
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    The question is right. I know DeMoivres, but in this case it's sin(x) + icos(x).

    • 2 years ago
  4. khurramshahzad Group Title
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    ah oookkk ya its absolute value

    • 2 years ago
  5. thrustfault Group Title
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    So far I have LHS = \[\frac{ e ^{ix}-e^{-ix} }{ 2 } + \frac{ e ^{ix}+e^{-ix} }{ 2 }i\] Have I completely overthought the problem ? :)

    • 2 years ago
  6. khurramshahzad Group Title
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    no i is denominator from demorvie`s, but in this case I did`nt get u how u got this result

    • 2 years ago
  7. khurramshahzad Group Title
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    ok am gonna solve its prove and implement according to this condition the `ll msg u here ok

    • 2 years ago
  8. thrustfault Group Title
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    Thanks :) I'm so confused...

    • 2 years ago
  9. khurramshahzad Group Title
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    hahahhah i just wanna say write i in denominator ok i mean (eix−e−ix)2+(eix+e−ix)2i

    • 2 years ago
  10. thrustfault Group Title
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    Hmm... I don't get it. How does that show that |sin(x) + icos(x)| = 1?

    • 2 years ago
  11. khurramshahzad Group Title
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    oh i see wait i solve it and post the diagrame okk buddy

    • 2 years ago
  12. thrustfault Group Title
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    Anyone?

    • 2 years ago
  13. khurramshahzad Group Title
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    daer i can`t .. i have seen the same result for it sqareroot 2 iz answer for both condition.. cos when we see cosx+isinx,, or sinx+icosx... taking coifficients we get the same result..

    • 2 years ago
  14. khurramshahzad Group Title
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    i thinks its wrong question or some thing missing ... can u give the link where did u got it from ???

    • 2 years ago
  15. thrustfault Group Title
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    Yep, it's question 1 iv) from this paper: http://www.scribd.com/doc/112148611/MATH1131-Mathematics-1A-1 Thanks :)

    • 2 years ago
  16. thrustfault Group Title
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    pdf attached here.

    • 2 years ago
  17. khurramshahzad Group Title
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    u got ur answer or not /??

    • 2 years ago
  18. thrustfault Group Title
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    Nope, heh. Not yet :(

    • 2 years ago
  19. khurramshahzad Group Title
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    ok if its difficault then msg me ur question i will solve it my self and will post the pic here ok

    • 2 years ago
  20. thrustfault Group Title
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    Yeah I don't know how to get the function to = 1...

    • 2 years ago
  21. AccessDenied Group Title
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    I am thinking that you are using the fact that |f(x)| represents the exact distance between it and 0 on the complex plane... Notice that you can create a right triangle with the real part of f(x) and the imaginary part of x on a complex plane: |dw:1352104120632:dw| So, here, Re(f) = sin(x) and Im(f) = cos(x). We drop off the i since we're dealing with distances only for the imaginary part. Using Pythagorean theorem, |f(x)|^2 = (Re(f))^2 + (Im(f))^2 = sin^2 x + cos^2 x. Since we have a neat trigonometric identity here, we can use it to reduce sin^2 x + cos^2 x as 1. Take the square root of both sides and we're done. :)

    • 2 years ago
  22. thrustfault Group Title
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    Thanks so much for your clear and concise answer!! I've been stuck on this question for a long time :)

    • 2 years ago
  23. AccessDenied Group Title
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    You're welcome! :)

    • 2 years ago
  24. sirm3d Group Title
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    perhaps this definition can refresh you \[\huge \left| a + bi \right|=a^2+b^2 \]

    • 2 years ago
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