Error using cos Not enough input arguments. HELP

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Hi all,
Can anyone help me with this problem - Error using cos Not enough input arguments.
x=0:0.25:5.*pi;
y=0:0.25:2.*pi;
[x,y]=meshgrid(x,y);
z = cos.*((x).*sin(y)).*((1-3.*cos((y).^2))-(x.^2).*sin((y).^2)).*cos(x);
Error using cos Not enough input arguments.
I'm new to MatLab and I need to plot a contour map using this function but I can't seem to find a way to get rid of the error message. Any help would be much appreciated.
Thanks, Kayn.

Risposta accettata

Azzi Abdelmalek
Azzi Abdelmalek il 6 Nov 2013
Modificato: Azzi Abdelmalek il 6 Nov 2013
You missed something in cos.* ,maybe you want
z = cos(x).*((x).*sin(y)).*((1-3.*cos((y).^2))-(x.^2).*sin((y).^2)).*cos(x);
You missed something in cos.
  4 Commenti
Kayn
Kayn il 6 Nov 2013
Yes, this works! Many, many thanks Azzi. I now have my contour maps and am trying to make them look nice. Thanks again.

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Più risposte (2)

Diego David Tubay Lozano
Diego David Tubay Lozano il 6 Mar 2021
diff(sin^2*([cos^2*(x^2)]))
Error using sin
Not enough input arguments.
help.
  1 Commento
Walter Roberson
Walter Roberson il 6 Mar 2021
syms x
diff(sin(x)^2 * cos(x^2)^2, x)
ans = 
or possibly you meant
syms x
diff(sin(cos(x^2)^2)^2, x)
ans = 

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Muhammad Syahmi Mohd Shamshul
5*cos(x)+4.2501-cos*(0.52367-x)
Error using cos
Not enough input arguments.
Error in Secant2 (line 13)
  1 Commento
Walter Roberson
Walter Roberson il 29 Giu 2021
You have a function named Secant2 that takes at least one parameter, that is named x in the function.
When you ran the function Secant2 you did not provide any parameters at the time of the call. You probably just pressed the big green Run button. When you press the green Run button, MATLAB will run the function without providing any parameters. If you run a function that has a variable (such as x) defined as a parameter, but you run the function without providing a value in that position, then MATLAB will never go hunting around elsewhere looking for an "x" defined somewhere to use as the value in the function. For example,
x = pi/7
y = sin
if sin were defined like
function sx = sin(x)
then MATLAB will never go looking in the workspace to find the pi/7 value to use for the function. If you want a function to use a parameter, then you have to pass the parameter to it.

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