put into exponential form

[gudi]

Put cos(x) cos(y) + i sin(x) sin(y) into exponential form. ( Only for
the special case when y is equal to pi/4, Euler relation can be
directly used, exp(i x)/ sqrt(2) )

Narasimham

[Dan Cass]

I guess you mean by exponential form r*e^(i*t).
Since that's r*cos(t) + i*r*sin(t), your form would satisfy
1) cos(x)cos(y) = r*cos(t)
2) sin(x)sin(y) = r*sin(t)
By squaring and adding, one can find r, which isn't
especially nice:
r = sqrt( [cos(x)*cos(y)]^2 + [sin(x)*sin(y)]^2 ).
Although one can say that tan(t) = tan(x)*tan(y),
it may not be as simple as
t = arctan(tan(x)*tan(y))...

[David W. Cantrell]

Dan Cass <dc...@sjfc.edu> wrote:
> > Put cos(x) cos(y) + i sin(x) sin(y) into exponential
> > form. ( Only for the special case when y is equal to
> > pi/4, Euler relation can be directly used, exp(i x)/ sqrt(2) )
> >
> > Narasimham
>
> I guess you mean by exponential form r*e^(i*t).
> Since that's r*cos(t) + i*r*sin(t), your form would satisfy
> 1) cos(x)cos(y) = r*cos(t)
> 2) sin(x)sin(y) = r*sin(t)
> By squaring and adding, one can find r, which isn't
> especially nice:
> r = sqrt( [cos(x)*cos(y)]^2 + [sin(x)*sin(y)]^2 ).

Alternatively, one could say

r = sqrt((1 + cos(2x)cos(2y))/2)

> Although one can say that tan(t) = tan(x)*tan(y),
> it may not be as simple as
> t = arctan(tan(x)*tan(y))...

True, it's not that simple. But I think we can say that

t = pi/2 sign(sin(x)sin(y)) - arctan(cot(x)cot(y))

at least when neither sin(x) nor sin(y) is zero. And I wouldn't be surprised if there were a nicer form for t, perhaps using the two-argument form of arctan.

David