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Question: How to verify set A and set B is equivalent?

Question:How to verify set A and set B is equivalent?

Aixleft math 120 Maple 2024
parameter set equivalence
March 08 2025
2

Hello, an interesting issue about set equivalence.

Logically, the two sets are equivalent by derivation.

The first principal case, for set A, k=0, the element values ​​are Pi/6 and 5*Pi/6, which corresponds to the case of k=0 (element value is Pi/6) and k=1 (element value is 5*Pi/6) in set B. Obviously, the k value is not one-to-one correspondence, but just a letter representing a positive integer. As for the second general case, it is the same steps I thought.

So, how to verify that the two sets are equivalent? I know Maple cannot do it in one step, but I don't know how to do it?

> 

A := solve(sin(x) = 1/2, allsolutions = true)

(1/6)*Pi+2*Pi*_Z5, (5/6)*Pi+2*Pi*_Z5

 (1)
> 

A := `assuming`([`union`({(1/6)*Pi+2*k*Pi}, {Pi-(1/6)*Pi+2*k*Pi})], [k::integer])

{(1/6)*Pi+2*k*Pi, (5/6)*Pi+2*k*Pi}

 (2)
> 

> 

B := `assuming`([{k*Pi+(1/6)*(-1)^k*Pi}], [k::integer])

{k*Pi+(1/6)*(-1)^k*Pi}

 (3)
> 

is(A = B)

false

 (4)
> 

restart
> 

alpha = 'alpha'

alpha = alpha

 (5)
> 

solve(sin(x) = alpha, x, allsolutions = true)

2*Pi*_Z1+arcsin(alpha), -arcsin(alpha)+Pi+2*Pi*_Z1

 (6)
> 

A := `assuming`([`union`({arcsin(alpha)+2*k*Pi}, {Pi-arcsin(alpha)+2*k*Pi})], [k::integer])

{arcsin(alpha)+2*k*Pi, Pi-arcsin(alpha)+2*k*Pi}

 (7)
> 

B := `assuming`([{k*Pi+(-1)^k*arcsin(alpha)}], [k::integer])

{k*Pi+(-1)^k*arcsin(alpha)}

 (8)
> 

is(A = B)

false

 (9)
 

NULL

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