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mathslover

  • one year ago

Are the angles \(\alpha\) and \(\beta\) given by, \(\large{\alpha = ( 2n + \frac{1}{2}) \pi \pm A}\) and \(\large{\beta = n \pi + (-1)^n (\frac{\pi}{2} - A)}\) same where \(\large{n \in I}\)

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  1. mathslover
    • one year ago
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    There is ^ belongs to " n belongs to I " I means integers

  2. mathslover
    • one year ago
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    @ParthKohli @experimentX @amistre64 @phi @.Sam.

  3. ParthKohli
    • one year ago
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    \[n \in \mathbb Z \]

  4. mathslover
    • one year ago
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    Yeah ... sorry I didn't know the sign for "belongs to" in LaTeX . btw any hint for the question ?

  5. mathslover
    • one year ago
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    I don't think they are equal.

  6. ParthKohli
    • one year ago
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    I'm looking at it. What is \(A\)?

  7. ParthKohli
    • one year ago
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    is \(A\) any real number?

  8. ParthKohli
    • one year ago
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    Of course it is.

  9. mathslover
    • one year ago
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    anything, that is not given in the question ..

  10. mathslover
    • one year ago
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    I tried to take : firstly n as even i.e. n = 2k and the second time n = 2k + 1

  11. mathslover
    • one year ago
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    I solved that for beta for i) n = 2k and then ii) n = 2k + 1

  12. ParthKohli
    • one year ago
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    Yes, that's what I have done too!

  13. mathslover
    • one year ago
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    that gives the resultant one as: \[\large{\beta = (2k + \frac{1}{2})\pi \pm A}\]

  14. ParthKohli
    • one year ago
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    \[\large \beta = n\pi + \dfrac{\pi}{2} - A = \dfrac{2\pi n+ \pi}{2}{\pm A}{}\]Yesh

  15. ParthKohli
    • one year ago
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    \[\large \alpha = \dfrac{4\pi n + \pi}{2} \pm A\]BTW, that's - A in the above post

  16. ParthKohli
    • one year ago
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    Your argument seems valid.

  17. ParthKohli
    • one year ago
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    gotta go, exam tomorrow. :-(

  18. mathslover
    • one year ago
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    well it should be solved like this : \[\large{\textbf{For n = 2k , n is even}}\] \[\large{k(2\pi) + \frac{\pi}{2} - A}\] \[\large{\textbf{ For n = 2k + 1, n is odd }}\] \[\large{(2k) \pi + \frac{\pi}{2} + A }\] \[\large{\beta = (2k)\pi + \frac{\pi}{2} \pm A}\] \[\large{\implies n\pi + \frac{\pi}{2}\pm A = \beta}\]

  19. mathslover
    • one year ago
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    therefore Beta is not equal to alpha ... Best of luck @ParthKohli

  20. mathslover
    • one year ago
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    @ash2326

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