[Maxima] how to show sin(30)=1/2, where 30 is a degree

[Leo Butler]

On Thu, 22 Oct 2009, 黃 裕雄 wrote:

< 1.how to show sin(30)=1/2, where 30 is a degree
< 
< we know if the angle is a radin then sin(30*%pi/180)=1/2,
< where 30 is a degree , How to  show sin(30)=1/2 in maxima ?

Maxima's sine function is defined in terms of radians. If you wish to
use degrees, then you can define your own function like so:

(%o197) sind(x):=sin(x*%pi/180)
(%i198) sind(30);
(%o198) 1/2

Generally, though, it's bad practice to measure angles in degrees when
doing mathematics.


< 
< (%i4) deg:30;
< (%o4)                                 30
< (%i5) sin(deg);
< (%o5)                               sin(30)
< (%i6) rad:deg*%pi/180;
<                                       %pi
< (%o6)                 !              ---
<                                        6
< (%i7) sin(rad);
<                                        1
< (%o7)                              -
<                                        2
< (%i8) (sin(rad))^! 2+(cos(rad))^2;
< (%o8)      &n! bsp;&nbs p;                          1
< (%i9) (sin(deg))^2+(cos(deg))^2;
<                                        2              2
< (%o9)                         sin (30) + cos (30)
< (%i10) (sin(x))^2+(cos(x))^2;
<                                             2         2
< (! %o10)                         sin (x) + cos (x)
< 
< 
< 2.how to show (sin(x))^2+(cos(x))^2=1 in maxima

Maxima knows this

(%i194) trigsimp(cos(x)^2 + sin(x)^2);
(%o194) 1



< 
< 
< 3. if sin(x)+cos(x)=1/2, solve sin(x)*cos(x)=?
< 
< we know (sin(x))^2+(cos(x))^2=1 ,
< so we have sin(x)*cos(x)=3/8.
< 
< Can it  solve in maxima ?
 
How about

(%i195) solve([x+y=1/2,x^2+y^2=1,z=x*y],[x,y,z]);
(%o195) [[x = (sqrt(7)+1)/4,y = -3/(2*sqrt(7)+2),z = -3/8],
         [x = -(sqrt(7)-1)/4,y = 3/(2*sqrt(7)-2),z = -3/8]]

Leo
-- 
The University of Edinburgh is a charitable body, registered in
Scotland, with registration number SC005336.