Re: sqrt(sin(x))+sqrt(cos(x)) = 1

From: Jeffrey A. Smith (kaladan_at_zianet.com)
Date: 11/05/04 

Date: Thu, 4 Nov 2004 19:57:18 -0700

That is how I would do it.

Your original equation is valid in the first quadrant (0, PI/2), so any
rotation that returns you to the same quadrant will give you a quartic of
the same form.

"Tassilo" <glutroteslicht@midwesternmail.com> wrote in message
news:be398bc3.0411040610.247e0de0@posting.google.com...
> "Jeffrey A. Smith" <kaladan@zianet.com> wrote in message
news:<4189ab1e$1@nntp.zianet.com>...
> > You might consider another approach that is a variation on Robin
Chapman's
> > post.
> >
> > Put u = sin x then cos x becomes sqrt (1 - u^2).
> > Then, after substitution and manipulation, you will find a quartic
equation
> > in u with two real roots (0 and 1) along with a quadratic root (with
> > imaginary roots).
> >
> > With the roots 0 and 1, you find that x becomes 0 or PI/2.
>
> Thanks for this good solution. But how to prove that the roots of this
> quartic are periodically repeating themselves ?
>
> By simply saying "sin and cos functions are periodic ones, so the
> roots are: k*2PI+(PI/2) and k*2PI, k element of N" ??
>
> Thanks again