Thank you!
On Friday, February 27, 2015 at 5:41:07 PM UTC+2, vdelecroix wrote:
>
> Here is one way... not sure it is the best
>
> sage: eq1 = sqrt(cos(4*x)+1)
> sage: eq2 = eq1.simplify_trig()
> sage: eq2
> sqrt(8*cos(x)^4 - 8*cos(x)^2 + 2)
>
> The next step consists in factoring what is insi
Here is one way... not sure it is the best
sage: eq1 = sqrt(cos(4*x)+1)
sage: eq2 = eq1.simplify_trig()
sage: eq2
sqrt(8*cos(x)^4 - 8*cos(x)^2 + 2)
The next step consists in factoring what is inside the sqrt:
sage: o = eq2.operands()[0]
sage: of = o.factor()
sage: o
8*cos(x)^4 - 8*cos(x)^2 + 2
s
OK. Let x is real.
How to rewrite sqrt(cos(4x)+1) into sqrt(2)abs(cos(2x))?
On Friday, February 27, 2015 at 3:36:59 PM UTC+2, Simon King wrote:
>
> Hi Paul,
>
> On 2015-02-27, Paul Royik > wrote:
> > What is the way to consistently simplify square roots of squares?
> >
> > Examples:
> >
> >
But...
sage: eq = sqrt((pi-5)^2)
sage: eq.canonicalize_radical()
pi - 5
And as you can read from the documentation
"""
Choose a canonical branch of the given expression. The square root,
cube root, natural log, etc. functions are multi-valued. The
"canonicalize_radical()" method will cho
Hi Paul,
On 2015-02-27, Paul Royik wrote:
> What is the way to consistently simplify square roots of squares?
>
> Examples:
>
> sqrt((x+1)^2) - > x+1
> sqrt(cos(4*x)+1) -> sqrt(2)cos(2x)
Simplification must not change the value of the expression. sqrt(x^2) is
certainly not equal to x. Even under
