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 2 years 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 \(\large{n \in I}\)
 2 years 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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mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0There is ^ belongs to " n belongs to I " I means integers

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0@ParthKohli @experimentX @amistre64 @phi @.Sam.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1\[n \in \mathbb Z \]

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

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0I don't think they are equal.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1I'm looking at it. What is \(A\)?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1is \(A\) any real number?

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0anything, that is not given in the question ..

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

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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, that's what I have done too!

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

ParthKohli
 2 years ago
Best 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

ParthKohli
 2 years ago
Best 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

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1Your argument seems valid.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1gotta go, exam tomorrow. :(

mathslover
 2 years ago
Best ResponseYou've already chosen the best response.0well 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
 2 years ago
Best ResponseYou've already chosen the best response.0therefore Beta is not equal to alpha ... Best of luck @ParthKohli
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