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thrustfault

  • 2 years ago

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

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  1. khurramshahzad
    • 2 years ago
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    hey its too much easy... |f(x)|=sqrt((1)^2+(1)^2)=sqrt(2)

  2. khurramshahzad
    • 2 years ago
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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)

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

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

  5. thrustfault
    • 2 years ago
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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 ? :)

  6. khurramshahzad
    • 2 years ago
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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

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

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

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

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

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

  12. thrustfault
    • 2 years ago
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    Anyone?

  13. khurramshahzad
    • 2 years ago
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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..

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

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

  16. thrustfault
    • 2 years ago
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    pdf attached here.

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

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

  19. khurramshahzad
    • 2 years ago
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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

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

  21. AccessDenied
    • 2 years ago
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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. :)

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

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

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

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