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Question: Fully simplify a constant containing `arctan`?

Question:Fully simplify a constant containing `arctan`?

sursumCorda 1284 Maple 2025
simplify simplification arctrig
5 hours ago
1 
f:=subs(r=(sqrt(5)+1)/2,7*arctan(r)^2+2*arctan(r^3)^2-arctan(r^5)^2);

The above expression is in fact equal to (7*Pi**2)/8 (which may be obtained by the `identify` command). 
As the following worksheet shows, simplify(f); fails to do full simplifications and the result consists of the less simplified arctan(2/11), while simplify(f/Pi^2); succeeds in making full simplifications and returning the remarkably simpler form: 

> 

restart;
> 

f := subs(r = (sqrt(5) + 1)/2, 7*arctan(r)^2 + 2*arctan(r^3)^2 - arctan(r^5)^2); # which is actually equal to “7/8*Pi^2”

7*arctan((1/2)*5^(1/2)+1/2)^2+2*arctan(((1/2)*5^(1/2)+1/2)^3)^2-arctan(((1/2)*5^(1/2)+1/2)^5)^2

 (1)
> 

is(f = 7/8*Pi^2);

true

 (2)
> 

simplify(f); # This cannot yield the desired result.

(11/16)*Pi^2+(3/4)*arctan(1/2)*Pi+(9/4)*arctan(1/2)^2-(1/4)*arctan(2/11)^2+(1/2)*arctan(2/11)*Pi

 (3)
> 

simplify(f/Pi^2); # To obtain the desired result, `f` has to be divided by Pi^2 manually.

7/8

 (4)
> 

simplify(arctan(2/11) - 3*arctan(2)); # So, Maple knows that arctan(2/11) = 3*arctan(2) - Pi (or, somewhat more elegantly, arccot(2/11) = 3*arccot(2)).

-Pi

 (5)
> 

simplify((3), constant); # No further simplification.

(11/16)*Pi^2+(3/4)*arctan(1/2)*Pi+(9/4)*arctan(1/2)^2-(1/4)*arctan(2/11)^2+(1/2)*arctan(2/11)*Pi

 (6)

As Maple does indeed know that arctan(2/11) = 3*arctan(2)-Pi, why can't Maple further simplify (3) by attempting to eliminate arctan(2/11) (so that we don't have to enter the identity manually)?

> 

simplify(simplify((3), {arctan(2/11) = 3*arctan(2) - Pi}));

(7/8)*Pi^2

 (7)
> 

simplify(subs(((sqrt(5)+1)*(1/2))^5 = ((sqrt(5)-1)*(1/2))*((sqrt(5)+1)*(1/2))^6, f))

(7/8)*Pi^2

 (8)
 

 

Download simplify_a_constant.mw

Since arctan(2/11) can be expressed by the significantly more concise arctan(1/2) and Pi (that is, arctan(2/11) = 3*arctan(2) - Pi), why didn't simplify(f); attempt to eliminate arctan(2/11) in the result to further simplify `f`?
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