ArcCosine function

this is the code i have so far. i need to write a function m-file that can solve the arc cosine function for [0, 2pi]. it is giving me an error when i try acosfull(0,0), it does not give me the error....what am i missing?

function [theta] = acosfull(x,y)
%acossfull: find output angle of arccos over 0<theta<2*pi
%   find value of acos(z) in the correct quadrant over the full range [0
%   2pi]

if y==x==0
    theta=0
    disp('Error.')
else r=sqrt(x^2 + y^2);
    z=x/r;
    acos(z)
end

[Victoria Walker]

One of my variables is a lowercase a, will this interfere?

[Walter Roberson]

Not at all. MATLAB does not have any implicit multiplication, not anywhere, so there is no danger that acos(z) might be interpreted as a*cos(z)

Answer 1

[Walter Roberson]

y==x==0

is parsed as

((y==x)==0)

So it compares x to y to give 0 (false) or 1 (true), and then it compares that 0 or 1 to 0. Effectively what you coded is

if y ~= x

Try

if y == x && y == 0

[Zelda Luxenburry]

perfect, thank you!!

Answer 2

[Wolfgang Garn]

In short theta = acos(x./sqrt(x.*x+y.*y)) + (y<0)*pi or below with a couple of tests.

function theta = acosfull(x,y)
%acossfull: find output angle of arccos over 0<theta<2*pi
%   find value of acos(z) in the correct quadrant over the full range [0 2pi]; 
%   acosfull requires as input x and y coordinates (length of x and y side of a rectangular triangle).
if nargin<1 % then give examples
  disp('A series of examples:');
  disp('45° = acosfull(2,2) radians: ');    acosfull(2,2)
  disp('90° = acosfull(0,3) radians: ');    acosfull(0,3)
  disp('135° = acosfull(-2,2) radians: ');    acosfull(-2,2)
  disp('180° = acosfull(-2,0) radians: ');    acosfull(-2,0)
  disp('225° = acosfull(-3,-3) radians: ');    acosfull(-3,-3)
  disp('270° = acosfull(0,-4) radians: ');    acosfull(0,-4)
  disp('315° = acosfull(2,-2) radians: ');    acosfull(2,-2)
  disp('360° = acosfull(2,-0.001) radians: ');    acosfull(2,-0.001)
  disp('NaN = acosfull(0,0): ');    acosfull(0,0)
  acosfull([2 -2 -2 -2 0], [2 2 -2 -2 0])
end
if nargin<2 % then display warning (and attempt original acos)
    warning('acosfull requires as input x and y coordinates (or length of x and y side of a rectangular triangle).');
    if nargin>0, theta = acos(x); end
else
        theta = acos(x./sqrt(x.*x+y.*y)) + (y<0)*pi;
end

[Mahmoud Taha]

basic trigonometry says cos is an even function and plot((1:-0.01:-1),acos(1:-0.01:-1)/pi) shows only positive output so all you need actually is:

(y>=0-y<0).*acos(x./sqrt(x.*x+y.*y))

I thought it is important to correct this misunderstanding

[Sai Zhou]

Nither of the answers works....

[Steven Lord]

Can you show us the code you ran and describe specifically what about the answers didn't work?

• Do you receive warning and/or error messages? If so the full and exact text of those messages (all the text displayed in orange and/or red in the Command Window) may be useful in determining what's going on and how to avoid the warning and/or error.
• Does it do something different than what you expected? If so, what did it do and what did you expect it to do?
• Did MATLAB crash? If so please send the crash log file (with a description of what you were running or doing in MATLAB when the crash occured) to Technical Support using the telephone icon in the upper-right corner of this page so we can investigate.

Answer 3

[Danay Rosh]

arccos(pi/2)

[madhan ravi]

arccos() is not even a valid syntax in MATLAB

[Walter Roberson]

arccos(pi/2) is valid syntax, but arccos() is not a function provided by MATLAB at the MATLAB level.

arccos is provided by the MuPAD programming language that runs inside the Symbolic Computing toolbox. https://www.mathworks.com/help/symbolic/mupad_ref/arccos.html . For example,

evalin(symengine, 'arccos(pi/2)')

is valid.

[madhan ravi]

Noted with care sir Walter.

[John D'Errico]

I might also add that the number acos(pi/2) will not even be a real number, but imaginary.

acos(pi/2)
ans =
            0 +     1.0232i

The set of inputs for which acos will produce REAL results is limited to the interval [-1,1].