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gerryliyana

  • 2 years ago

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  1. gerryliyana
    • 2 years ago
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    Show that: \[|cosz|^{2}= \cos^{2}x+\sinh^{2}x\] Hint: using \[\cos z = \frac{ e^{iz} + e^{-iz}}{ 2 }\] \[\sin z = \frac{ e^{iz} - e ^{-iz} }{ 2i }\] \[\cosh z = \frac{ e^{z}+e^{-iz} }{ 2 }\] \[\sinh z = \frac{ e^{z} - e^{-z} }{ 2 }\]

  2. gerryliyana
    • 2 years ago
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    Sorry I MEAN, show that \[|\cos z|^{2} = \cos^{2} x + \sinh^{2}y\]

  3. gerryliyana
    • 2 years ago
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    would you kindly help me again @niksva ??

  4. niksva
    • 2 years ago
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    Take LHS first (cosz)^2 replace z with x +iy cos(x+iy) = cos(x)cos(iy) - sin(x)sin(iy)

  5. niksva
    • 2 years ago
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    as cos(iy) = coshy and sin(iy) =i sinhy therefore cosz = cos(x)cos(hy)-i sin(x)sin(hy)

  6. niksva
    • 2 years ago
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    @gerryliyana after squaring and applying (a-b)^2 formula , the equation become difficult to solve

  7. gerryliyana
    • 2 years ago
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    i've tried.., but didn't match,

  8. niksva
    • 2 years ago
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    how did u approach the problem?

  9. gerryliyana
    • 2 years ago
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    using hint above..., i tried, one by one.., hbu?? it's make sense ?

  10. niksva
    • 2 years ago
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    how can u apply above hint in RHS?

  11. gerryliyana
    • 2 years ago
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    what the next ??

  12. niksva
    • 2 years ago
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    my ques is as z = x + iy where x is real number and y is imaginary number can we write cosx= (e^ix+ e^-ix)/2 ???????/ according to me, we cannot write this as x is a real number

  13. niksva
    • 2 years ago
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    @UnkleRhaukus help?

  14. electrokid
    • 2 years ago
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    \[\Large{ \cos z={e^{-iz}+e^{iz}\over2}\\ LHS=|\cos z|^2={1\over4}\left(e^{-2iz}+2e^{-iz+iz}+e^{2iz}\right)\\ \quad={1\over2}+{1\over4}\left[e^{-2i(x+iy)}+e^{2i(x+iy)}\right]\\ \quad=\color{red}{{1\over4}\left(2+e^{2y}+e^{-2y}\right)+{1\over4}\left(e^{-i2x}+e^{i2x}\right)}\\ RHS=\cos^2x+\sinh^2y\\ \quad={1\over4}\left(e^{-i2x}+e^{i2x}+2\right)-{1\over4}\left(e^{2y}-2+e^{-2y}\right)\\ }\] They do not seem to match!!

  15. gerryliyana
    • 2 years ago
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    have idea ???

  16. electrokid
    • 2 years ago
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    yes. you can prove it thus way: 1) get LHS = "a" 2) if you can show RHS = "a", then LHS=RHS="a" and QED. but following this, the two sides probably have an issue with the "signs"

  17. electrokid
    • 2 years ago
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    no wait.. my bad.. I took "sin y" instead of "sinh y" \[\Large{ \cos z={e^{-iz}+e^{iz}\over2}\\ LHS=|\cos z|^2={1\over4}\left(e^{-2iz}+2e^{-iz+iz}+e^{2iz}\right)\\ \quad={1\over2}+{1\over4}\left[e^{-2i(x+iy)}+e^{2i(x+iy)}\right]\\ \quad=\color{red}{{1\over4}\left(2+e^{2y}+e^{-2y}\right)+{1\over4}\left(e^{-i2x}+e^{i2x}\right)}\\ RHS=\cos^2x+\sinh^2y\\ \quad={1\over4}\left(e^{-i2x}+e^{i2x}+2\right)+{1\over4}\left(e^{2y}-2+e^{-2y}\right)\\ \quad=\color{red}{{1\over4}\left(e^{2y}+e^{-2y}\right)+{1\over4}\left(e^{-i2x}+e^{i2x}\right)}\\ {\rm SO,}\\ \color{green}{\boxed{|\cos z|^2={1\over2}+\cos^2x+\sinh^2y}} }\]

  18. gerryliyana
    • 2 years ago
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    but it's not the answer

  19. electrokid
    • 2 years ago
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    as you see, the LHS and the RHS differ by \(1\over2\).

  20. gerryliyana
    • 2 years ago
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    yes., i see, did you know why ???

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