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A procedure to find tan inverse in PL/SQL

From: Mahesh Vallampati <m0v5533_at_tam2000.tamu.edu>
Date: 1995/06/12
Message-ID: <3rhvrn$qr3_at_news.tamu.edu>#1/1 







/* 


This is a program which will calculate the tan inverse of a value and
give the angle in radians. The input to the procedure taninverse is
taninvx which is the value for which we have to find the angle. The
output obtained is the angle which is in radians. The infinite series
approximation for taninv(x) has been used to find the angle. The series
has been approximated upto n terms depending on where taninvx lies.
The angle returned is in the region ( 0,PI ).
*/


-- Procedure which will calculate the taninverse of a value and give the




    
  
  •   angle in radians.
/* 
	Author : Mahesh Vallampati
	Department of Electrical Engineering
	Texas A and M University, College Station, Texas 77840.
Important: The software and the program is provided as is and you are using 
it at your own risk. 
In case you find it useful, please send me an e-mail to oracle7_at_tamu.edu.
If you have any corrections/suggestions please send them to the same address.
*/
/* 
This can also be used to find arcsin,arccos. The  procedure is outlined as
follows. Given sin(theta) where theta is the angle to be found, use the well
known trignometric identity 
cos(theta) = sqrt(1-(sin(theta)*sin(theta))) 
and hence obtain 
tan(theta)=sin(theta)/cos(theta) 
and then use the procedure below to find the angle. 
Similarly cos(theta) where theta is the angle to be found can be done as 
sin(theta)=sqrt(1-(cos(theta)*cos(theta))) 
and hence obtain
tantheta=sin(theta)/cos(theta) 
and then use the procedure.
*/
/* KNOWN BUGS 

For large values of taninvx the Oracle error: ORA-06502 occurs.
ORA-06502 PL/SQL : numeric or value error.
This error is happening becase the power function is being used. For large
values of x or y  in power(x,y) the value exceeds the range and hence causes
error
*/







declare


tantheta1 real;


newtheta1 real;




    
  
  •   Procedure to find taninverse begins here
procedure taninverse(radian IN OUT real,taninvx IN real) IS
M_PI_2 CONSTANT REAL :=1.57079632679489661923;

M_PI_4 CONSTANT REAL :=0.78539816339744830962;

i number(4);
j number(4);
k number(4);
n number(4);
p number(6);
temp real;
begin
k:=1;
n:=120;
if taninvx = 0.0 then
radian:=0.0;
else
if abs(taninvx) > 1
then
	radian:=M_PI_2;
		if abs(taninvx) > 11 
		then
			if abs(taninvx) < 20
			then
				n:=60;
			else
				n:=30;
			end if;
		end if;
		for i in 1..n
		loop
			if mod(i,2) > 0
			then
				temp:=power(-1,k)/(i*(power(taninvx,i)));
				radian:=radian+temp;		
				k:=k+1;
			end if;
		end loop;
elsif ( abs(taninvx) ) < 1
then
	radian:=0.0;
		for i in 1..n
		loop
			if mod(i,2) > 0
			then
				temp:=power(-1,k+1)*(power(taninvx,i))/i;
				radian:=radian+temp;		
				k:=k+1;
			end if;
		end loop;
elsif taninvx = 1 
then
	radian:= M_PI_4;
elsif taninvx = -1 
then
	radian:= M_PI_4 + M_PI_2 ;
end if;





	if radian < 0
	then
		radian:= radian +  2*M_PI_2;
	end if;




end if;


end  taninverse;



    
  
  •   Procedure to find taninverse ends here






begin


tantheta1:=	; /* A value for which you want to find the tangent */
taninverse(newtheta1,tantheta1);


-- The angle is stored in newtheta1 n radians and can be used


end;


/






Thanks


Mahesh Vallampati


oracle7_at_tamu.edu


If anybody can test the procedure and comment upon it, I would appreciate it.
Received on Mon Jun 12 1995 - 00:00:00 CEST












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