## mathslover 3 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 $$\large{n \in I}$$

1. mathslover

There is ^ belongs to " n belongs to I " I means integers

2. mathslover

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

3. ParthKohli

$n \in \mathbb Z$

4. mathslover

Yeah ... sorry I didn't know the sign for "belongs to" in LaTeX . btw any hint for the question ?

5. mathslover

I don't think they are equal.

6. ParthKohli

I'm looking at it. What is $$A$$?

7. ParthKohli

is $$A$$ any real number?

8. ParthKohli

Of course it is.

9. mathslover

anything, that is not given in the question ..

10. mathslover

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

11. mathslover

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

12. ParthKohli

Yes, that's what I have done too!

13. mathslover

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

14. ParthKohli

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

15. ParthKohli

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

16. ParthKohli

17. ParthKohli

gotta go, exam tomorrow. :-(

18. mathslover

well it should be solved like this : $\large{\textbf{For n = 2k , n is even}}$ $\large{k(2\pi) + \frac{\pi}{2} - A}$ $\large{\textbf{ For n = 2k + 1, n is odd }}$ $\large{(2k) \pi + \frac{\pi}{2} + A }$ $\large{\beta = (2k)\pi + \frac{\pi}{2} \pm A}$ $\large{\implies n\pi + \frac{\pi}{2}\pm A = \beta}$

19. mathslover

therefore Beta is not equal to alpha ... Best of luck @ParthKohli

20. mathslover

@ash2326