Problem with is and equal testing equality of expressions

Dear All --

 I tried using this "is" operator today, and I failed miserably:

(%i27) is(equal ((x + 1)^2, x^2 + 2*x + 1));
(%o27)                               true

 But...

(%i28) is(equal(sin(a+b),sin(a)*cos(b)+cos(a)*sin(b)));
Maxima was unable to evaluate the predicate:
equal(sin(b + a), cos(a) sin(b) + sin(a) cos(b))
 -- an error.  Quitting.  To debug this try debugmode(true);
(%i29)

 I don't grok the debugger, so I've got no usefull output.

(%i18) is((cos(t))**2=(1+cos(2*t))/2);
(%o18)                               false
(%i19) is(sin(a+b)=sin(a)*cos(b)+cos(a)*sin(b));
(%o19)                               false

 Is it me or is Maxima?
 Here's what the nice documentation little man inside my computer said:

(%i9) describe(equal)
;

 0: (maxima.info)Inequality.
 1: equal :Definitions for Operators.
 2: equalp :Definitions for Fourier Series.
 3: notequal :Definitions for Operators.
 4: setequalp :Definitions for Sets.
Enter space-separated numbers, `all' or `none': 1

 -- Function: equal (<expr_1>, <expr_2>)
     Used with an `is', returns `true' (or `false') if and only if
     <expr_1> and <expr_2> are equal (or not equal) for all possible
     values of their variables (as determined by `ratsimp').  Thus `is
     (equal ((x + 1)^2, x^2 + 2*x + 1))' returns `true' whereas if `x'
     is unbound `is ((x + 1)^2 = x^2 + 2*x + 1)' returns `false'.  Note
     also that `is(rat(0)=0)' yields `false' but `is (equal (rat(0),
     0))' yields `true'.

     If a determination can't be made, then `is (equal (a, b))' returns
     a simplified but equivalent expression, whereas `is (a=b)' always
     returns either `true' or `false'.

     All variables occurring in <expr_1> and <expr_2> are presumed to
     be real valued.

     The negation of `equal' is `notequal'.  Note that because of the
     rules for evaluation of predicate expressions (in particular
     because `not <expr>' causes evaluation of <expr>), `notequal' is
     not equivalent to `not equal' in some cases.

     `ev (<expr>, pred)' is equivalent to `is (<expr>)'.

          (%i1) is (x^2 >= 2*x - 1);
          (%o1)                         true
          (%i2) assume (a > 1);
          (%o2)                        [a > 1]
          (%i3) is (log (log (a+1) + 1) > 0 and a^2 + 1 > 2*a);
          (%o3)                         true


  Completely bewildered, gratefull in advance for any help,
  regs,

   Henry Lenzi