There is a variety of functions that can change the form of trigonometric
expressions. You could try to understand in detail the strengths and
weaknesses of each function, or you can do what I do, which is to try
various approaches until one gets you the form you're looking for:
(%i8) e1: (sin(2*a) + sin(5*a) - sin(3*a)) / (cos(a) + 1 - 2*sin(2*a)^2);
(%o8) (sin(5*a)-sin(3*a)+sin(2*a))/(-2*sin(2*a)^2+cos(a)+1)
(%i9) factor(trigexpand(e1));
(%o9)
-sin(a)*(sin(a)^4-10*cos(a)^2*sin(a)^2+sin(a)^2+5*cos(a)^4-3*cos(a)^2+2*cos(a))/(8*cos(a)^2*sin(a)^2-cos(a)-1)
(%i10) factor(trigsimp(e1));
(%o10) -(sin(5*a)-sin(3*a)+sin(2*a))/(2*sin(2*a)^2-cos(a)-1)
(%i11) factor(trigreduce(e1));
(%o11) (sin(5*a)-sin(3*a)+sin(2*a))/(cos(4*a)+cos(a))
(%i12) factor(trigrat(e1));
(%o12) 2*sin(a)
Your first example:
(%i14) e2: (sin(a)^6 + cos(a)^6 + 3*sin(a)^2 * cos(a)^2);
(%o14) sin(a)^6+3*cos(a)^2*sin(a)^2+cos(a)^6
(%i15) trigexpand(e2);
(%o15) sin(a)^6+3*cos(a)^2*sin(a)^2+cos(a)^6
(%i16) factorsum(trigexpand(e2));
(%o16) sin(a)^6+3*cos(a)^2*sin(a)^2+cos(a)^6
(%i17) factorsum(trigreduce(e2));
(%o17) 1
(%i18) factorsum(trigsimp(e2));
(%o18) 1
(%i19) factorsum(trigrat(e2));
(%o19) 1
When you suspect the expression may simplify to a constant, you might even
try 'taylor':
(%i20) taylor(e2,a,0,20);
(%o20) +1
Another approach is exponentialization. You'd do something like
rectform(radcan(exponentialize( ... ))) or rectform(factor(exponentialize(
... ))). This in fact gets e2 => 1, but doesn't help much with e1.
Does this resolve your issue?
-s
On Wed, Oct 19, 2011 at 13:13, Jalil Karimov <jukarimov at gmail.com> wrote:
> (sin(2*a) + sin(5*a) - sin(3*a)) / (cos(a) + 1 - 2*sin(2*a)^2)