Question

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 \(\large{n \in I}\)

Answer 1

There is ^ belongs to

" n belongs to I " I means integers

Answer 2

@ParthKohli @experimentX @amistre64 @phi @.Sam.

Answer 3

\[n \in \mathbb Z \]

Answer 4

Yeah ... sorry I didn't know the sign for "belongs to" in LaTeX .

btw any hint for the question ?

Answer 5

I don't think they are equal.

Answer 6

I'm looking at it.

What is \(A\)?

Answer 7

is \(A\) any real number?

Answer 8

Of course it is.

Answer 9

anything, that is not given in the question ..

Answer 10

I tried to take : firstly n as even i.e. n = 2k and the second time n = 2k + 1

Answer 11

I solved that for beta for i) n = 2k and then ii) n = 2k + 1

Answer 12

Yes, that's what I have done too!

Answer 13

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

Answer 14

\[\large \beta = n\pi + \dfrac{\pi}{2} - A = \dfrac{2\pi n+ \pi}{2}{\pm A}{}\]Yesh

Answer 15

\[\large \alpha = \dfrac{4\pi n + \pi}{2} \pm A\]BTW, that's - A in the above post

Answer 16

Your argument seems valid.

Answer 17

gotta go, exam tomorrow. :-(

Answer 18

well it should be solved like this :

\[\large{\textbf{For n = 2k , n is even}}\]
\[\large{k(2\pi) + \frac{\pi}{2} - A}\]
\[\large{\textbf{ For n = 2k + 1, n is odd }}\]
\[\large{(2k) \pi + \frac{\pi}{2} + A }\]
\[\large{\beta = (2k)\pi + \frac{\pi}{2} \pm A}\]
\[\large{\implies n\pi + \frac{\pi}{2}\pm A = \beta}\]

Answer 19

therefore Beta is not equal to alpha ...

Best of luck @ParthKohli

Answer 20

@ash2326