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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 \(\large{n \in I}\)
 one year ago
 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 \(\large{n \in I}\)
 one year ago
 one year ago

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mathsloverBest ResponseYou've already chosen the best response.0
There is ^ belongs to " n belongs to I " I means integers
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
@ParthKohli @experimentX @amistre64 @phi @.Sam.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
\[n \in \mathbb Z \]
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
Yeah ... sorry I didn't know the sign for "belongs to" in LaTeX . btw any hint for the question ?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
I don't think they are equal.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
I'm looking at it. What is \(A\)?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
is \(A\) any real number?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
anything, that is not given in the question ..
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
I tried to take : firstly n as even i.e. n = 2k and the second time n = 2k + 1
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
I solved that for beta for i) n = 2k and then ii) n = 2k + 1
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Yes, that's what I have done too!
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
that gives the resultant one as: \[\large{\beta = (2k + \frac{1}{2})\pi \pm A}\]
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
\[\large \beta = n\pi + \dfrac{\pi}{2}  A = \dfrac{2\pi n+ \pi}{2}{\pm A}{}\]Yesh
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
\[\large \alpha = \dfrac{4\pi n + \pi}{2} \pm A\]BTW, that's  A in the above post
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
Your argument seems valid.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.1
gotta go, exam tomorrow. :(
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
well it should be solved like this : \[\large{\textbf{For n = 2k , n is even}}\] \[\large{k(2\pi) + \frac{\pi}{2}  A}\] \[\large{\textbf{ For n = 2k + 1, n is odd }}\] \[\large{(2k) \pi + \frac{\pi}{2} + A }\] \[\large{\beta = (2k)\pi + \frac{\pi}{2} \pm A}\] \[\large{\implies n\pi + \frac{\pi}{2}\pm A = \beta}\]
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
therefore Beta is not equal to alpha ... Best of luck @ParthKohli
 one year ago
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