trigsimp() would be what would do it, if anything.

Some recent changes (after the last release) have improved trigsimp,
so that it can now do this.  As a work-around, you can just factor it
first

In [11]: print trigsimp(factor(expr2))
1

Another trick that works well is to replace sin(x)**2 with 1 - cos(x)**2:

In [15]: expr2.subs(sin(s)**2, 1 - cos(s)**2).expand()
Out[15]: 1

Or you could try the git master trigsimp, which is much more powerful
than the one in the release.  See
https://github.com/sympy/sympy/wiki/Getting-the-bleeding-edge.

Aaron Meurer

On Wed, Jan 23, 2013 at 2:10 PM, Bryan Jones <[email protected]> wrote:
> All,
>
> I'd like SymPy to simplify some trig expressions (see below), but I'm
> stumped. What's the appropriate call (I assume to one of the functions in
> http://docs.sympy.org/0.7.2/modules/simplify/simplify.html)? I've tried
> several and can't seem to figure out the right approach.
>
>
> from sympy import symbols, sin, cos
>
> from sympy.simplify import *
>
> s = symbols('s')
>
> expr = sin(s)**2 + cos(s)**2
>
> print('Original expression: %s' % expr)
>
> print('Simplifies to: %s' % simplify(expr))
>
> expr2 = (expr**2).expand()
>
> print('Original squared: %s' % expr2)
>
> print('Simplifies to: %s' % trigsimp(expr2))
>
>
> The output:
>
> Original expression: sin(s)**2 + cos(s)**2
>
> Simplifies to: 1
>
> Original squared: sin(s)**4 + 2*sin(s)**2*cos(s)**2 + cos(s)**4
>
> Simplifies to: sin(s)**4 - cos(s)**4 + 2*cos(s)**2
>
>
> I'd like the last line to read: "Simplifies to: 1"
>
> Any ideas?
>
> Thanks!
>
> Bryan
>
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