MATLAB: Matlab Trigonometry and built in functions

MATLABtrigonometry

Hello Dear Forum Members;

>> cos(pi/5)*cos(pi/5)=(tan(pi/5)+sin(pi/5))/(2*tan(pi/5))
 cos(pi/5)*cos(pi/5)=(tan(pi/5)+sin(pi/5))/(2*tan(pi/5))
                    ↑
Error: The expression to the left of the equals sign is not a valid target for an assignment.
>> cos(pi/5)*cos(pi/5)
ans =
      0.6545

I have written the codes above but they are not matched. I kindly request your help. Thank you…

Best Answer

You are missing the /2 inside the cosine on the lhs. E.g.,

>> cos((pi/5)/2)^2
ans =
   0.904508497187474

Related Solutions

MATLAB: Solving non linear equation using fsolve

You are not passing theta_n and other parameters to root2d(). One fix is to make root2d a Nested Function. Also, you must choose a different initial point. phi0 = [10,10] worked for me.

function solve_it()
    beta_a = 10;
    alpha_n = 10 ;
    eta= 18;
    i = 18;
    theta_i = 10;
    theta_n = 10 ;
    
    fun = @root2d;
    phi0 = [10,10];
    phi = fsolve(fun,phi0);
    phi_n = phi(1),
    phi_i = phi(2),
    function F = root2d(phi)
    F = zeros(2,1);
    F(1) = (pi*sin((pi*(phi(1) + theta_n))/180)*cos((pi*phi(2))/180)*(cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180)))/(180*sin((pi*phi(1))/180)*(cos((pi*(phi(1) + theta_n))/180)*cos((phi(2)*pi)/180) + sin((phi(2)*pi)/180)*tan((theta_i*pi)/180))^2) - (pi*cos((pi*phi(1))/180)*(cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180)))/(180*sin((phi(1)*pi)/180)^2*(cos((pi*(phi(1) + theta_n))/180)*cos((pi*phi(2))/180) + sin((pi*phi(2))/180)*tan((pi*theta_i)/180)));
    F(2) = ((cos((pi*theta_n)/180) + tan((pi*i)/180)*tan((pi*theta_i)/180))*((pi*cos((pi*(phi(1) + theta_n))/180)*sin((pi*phi(2))/180))/180 - (pi*cos((pi*phi(2))/180)*tan((pi*theta_i)/180))/180))/(sin((pi*phi(1))/180)*(cos((pi*(phi(1) + theta_n))/180)*cos((phi(2)*pi)/180) + sin((phi(2)*pi)/180)*tan((theta_i*pi)/180))^2);
    end
end

MATLAB: Trigonometric equation thestery

First, you need to understand that sqrt(3)/2 is NOT 0.8660. There is a difference. Small errors like that will often be significant.

Second, you need to recognize that sin and cos are written to work in RADIANS. This is true of most programming languages. So, while 60 may mean 60 degrees to you, MATLAB will generally interpret it in the form of 60 radians.

So, IF 60 is intended to be in the form of radians, then we have:

sin(60)
ans =
     -0.30481
cos(60)
ans =
     -0.95241

However, since you seem to have substituted 1/2 for cos(60), I can only assume that you intended the number 60 to be in degrees.

Substituting the identity you used, we get:

soln = solve(2*cos(2*x) - 2*(cos(-2*x)*1/2 - sin(-2*x)*sqrt(3)/2) == 0,x)
soln =
 -(log(- 3^(1/2)/2 - 1i/2)*1i)/2
   -(log(3^(1/2)/2 + 1i/2)*1i)/2

That may not look like 15 degrees to you, but...

rad2deg(double(soln))
ans =
          -75 + 1.6214e-71i
           15 + 1.6214e-71i

Feel free to ignore the imaginary part, that is 71 powers of 10 smaller than the real part. It is zero.

So there is no mystery at all. Apparently the symbolic toolbox was intelligent enough that it saw your original equation with the number 60 inside a trig function, and realized that you must be working in degrees.

Best Answer

Related Solutions

MATLAB: Solving non linear equation using fsolve

MATLAB: Trigonometric equation thestery

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