trigcos() function

[DrD]

s:=√((sin(t))²+(-cos(t)+1)²)
k:=expand(s) ---> √(-2*cos(t)+(cos(t))²+1+(sin(t))²)

trigcos(√(-2*cos(t)+(cos(t))²+1+(sin(t))²)) ---> √(-2.*cos(t)+2.)
trigcos(k) ---> 2.*ABS(sin(0.5*t))

With both trigcos() cases the same, why the additional simplification on trigcos(k)?

How can I get the trigcos(k) result: √(-2.*cos(t)+2.)?

Thanks,

-Dale-

[parisse]

Maybe because you are not in exact mode.