Why does matlab drop the negative sign when evaluating symbolic derivative of x(t) = cos(Wn*t+phi)?

I wrote a short script to take the derivatives of some symbolic functions to check my math, however I get a slightly different result than when I take the derivatives by hand, and I believe my hand results are correct. Perhaps MATLAB is handling one of my symbols (possibly phi) or the function in a way I don't understand as a relative newb to MATLAB.

Bottom line, it seems to be dropping the negative sign when evaluating x'(t), where x=Acos(Wn*t-phi). When I do it by hand I get x'(t)=-A*Wn*sin(Wn*t-phi), (because derivative of cosine is negative sine, and pulling out the Wn by the chain rule), but when MATLAB does diff(x,t) it does not give the negative sign, just x'(t)=A*Wn*sin(Wn*t-phi). Note the negative in front of the A is not there.

I'm guessing it has something to do with how I'm defining the function, the symbol phi or Wn, or maybe I have something else wrong, but if someone has some insight, I would love to hear what it is so I can understand what's going on and do this more reliably and with more trust in the result in the future. Code below for reference. Thanks!

syms A zunderdamped zoverdamped Xo Vo Wn Wd t phi;

xundamped = A*cos((Wn*t)-phi);
diffxundamped = diff(xundamped,t)
%this gives the result: diffxundamped = A*Wn*sin(phi - Wn*t)
%which is incorrect,     actual x'(t) = -A*Wn*sin(Wn*t-phi)

xunderdamped = A*(exp(-zunderdamped*Wn*t))*cos(Wd*t-phi);
diffxunderdamped = diff(xunderdamped, t)
%this gives diffxunderdamped = A*Wd*exp(-Wn*t*zunderdamped)*sin(phi - Wd*t) - A*Wn*zunderdamped*cos(phi - Wd*t)*exp(-Wn*t*zunderdamped)
%which is also incorrect, it dropped the negative again, actual x'(t) = -A*Wd*exp(-Wn*t*zunderdamped)*sin(phi - Wd*t) - A*Wn*zunderdamped*cos(phi - Wd*t)*exp(-Wn*t*zunderdamped)

xcriticallydamped = (Xo+(Vo+Wn*Xo)*t)*exp(-Wn*t);
diffxcriticallydamped = diff(xcriticallydamped,t)
%this gives diffxcriticallydamped = exp(-Wn*t)*(Vo + Wn*Xo) - Wn*exp(-Wn*t)*(Xo + t*(Vo + Wn*Xo))
%which is correct (no derivative of cosine evaluated in this one though)

Answer 1

[Walter Roberson]

sin(-x) = -sin(x) so A*Wn*sin(phi - Wn*t) is the same as -A*Wn*sin(Wn*t-phi) -- notice that the sign of the inside of the sin() is negated.

[JustAnotherHumanoid]

Yes! Of course, I can't believe I missed that. You're the man! Thanks very much.