mathslover
  • mathslover
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 \(\large{n \in I}\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathslover
  • mathslover
There is ^ belongs to " n belongs to I " I means integers
mathslover
  • mathslover
@ParthKohli @experimentX @amistre64 @phi @.Sam.
ParthKohli
  • ParthKohli
\[n \in \mathbb Z \]

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mathslover
  • mathslover
Yeah ... sorry I didn't know the sign for "belongs to" in LaTeX . btw any hint for the question ?
mathslover
  • mathslover
I don't think they are equal.
ParthKohli
  • ParthKohli
I'm looking at it. What is \(A\)?
ParthKohli
  • ParthKohli
is \(A\) any real number?
ParthKohli
  • ParthKohli
Of course it is.
mathslover
  • mathslover
anything, that is not given in the question ..
mathslover
  • mathslover
I tried to take : firstly n as even i.e. n = 2k and the second time n = 2k + 1
mathslover
  • mathslover
I solved that for beta for i) n = 2k and then ii) n = 2k + 1
ParthKohli
  • ParthKohli
Yes, that's what I have done too!
mathslover
  • mathslover
that gives the resultant one as: \[\large{\beta = (2k + \frac{1}{2})\pi \pm A}\]
ParthKohli
  • ParthKohli
\[\large \beta = n\pi + \dfrac{\pi}{2} - A = \dfrac{2\pi n+ \pi}{2}{\pm A}{}\]Yesh
ParthKohli
  • ParthKohli
\[\large \alpha = \dfrac{4\pi n + \pi}{2} \pm A\]BTW, that's - A in the above post
ParthKohli
  • ParthKohli
Your argument seems valid.
ParthKohli
  • ParthKohli
gotta go, exam tomorrow. :-(
mathslover
  • mathslover
well it should be solved like this : \[\large{\textbf{For n = 2k , n is even}}\] \[\large{k(2\pi) + \frac{\pi}{2} - A}\] \[\large{\textbf{ For n = 2k + 1, n is odd }}\] \[\large{(2k) \pi + \frac{\pi}{2} + A }\] \[\large{\beta = (2k)\pi + \frac{\pi}{2} \pm A}\] \[\large{\implies n\pi + \frac{\pi}{2}\pm A = \beta}\]
mathslover
  • mathslover
therefore Beta is not equal to alpha ... Best of luck @ParthKohli
mathslover
  • mathslover
@ash2326

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