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

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

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

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

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

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

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

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

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

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

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