Reducing a signal to Acos(omega * t+ phi) with negative omega*t I am trying to solve a phasor addition problem, reducing from its original form to 
$$X(t) = A \cos(\omega_0 * t + \phi)$$

The original equation is :
 $$X(t) = 2 \sin(\omega_0 * t + 45) + \cos(\omega_0*t)$$

I converted the sine term into cosine. The result that i got was 
$$X(t) = 2 \cos(\pi/2 - \omega_0*t - \pi/4) + \cos(\omega_0*t) $$
Which reduces to :
$$X(t) = 2 \cos(\pi/4 - \omega_0*t) + \cos(\omega_0*t) $$
Up to this point I verify in MATLAB that the original function and the function in cosines are equal by graphing them and then calling the $norm()$ function and verify that it is something very close to 0. 
http://www.mathworks.com/help/matlab/ref/norm.html

I then perform the phasor addition process. 
$X(t) = Re(2e^{j(\pi/4)}e^{j\omega_0 t})+Re(e^{j0}e^{j\omega_0 t})$ 

$$X_1= 2e^{j(\pi/4)} = 2(cos(\pi/4) + jsin(\pi/4)) = 2(\frac{\sqrt2}2+j\frac{\sqrt2}2) = \sqrt2 + j\sqrt2$$
$$X_2 = e^{j0} = cos0 + jsin0 = 1$$
$$X_3 = X_1 + X_2 = \sqrt2 + j\sqrt2 + 1 = 2.41421 + j1.41421$$
 To turn it into back into the stated form of $X(t) = A \cos(\omega_0 * t + \phi)$ i find the $R$ 
$$R = \sqrt{2.41421^2 + 1.41421^2} = 2.79793$$
$$\theta = arctan(1.41421/2.41421) = .529902$$

The final result being $X(t) = 2.79793\cos(\omega_0 t + .529902) $

I then go to MATLAB to plot the original signal and the reduced signal that I found but when I call the $norm(originalSignal - reducedSignal)$ I get a number much greater than 0.

I am trying to figure out why this is returning such a high number. Some of my guesses are the negative $-\omega_0 t$ in the first term of the function when I convert sine to cosine $X(t) = 2 \cos(\pi/2 - \omega_0 t - \pi/4) + \cos(\omega_0 t) $ I don't know if the process changes for a situation like that. Problems I have worked through and have gotten the correct answer always seem to have the $\omega_0 t$ as positive part inside the cosine term.

Below is the Matlab code.

t = [0 : .01 : 1];
originalWave = 2*sin(50 *t + pi/4) + cos(50*t);
reducedWave = 2.79793*cos(50*t+.529902);

subplot(2,1,1)
plot(t, originalWave)
subplot(2,1,2)
plot(t, reducedWave)

norm(originalWave - reducedWave)


Any ideas on what could be wrong with my process or answer?

I appreciate any advice or help. Thanks.
You flipped a sign on the phase, the answer is -.529....
It's not 
$$X(t) = Re(2e^{j(\pi/4)}e^{j\omega_0 t})+Re(e^{j0}e^{j\omega_0 t})$$

It's
$$X(t) = Re(2e^{-j(\pi/4)}e^{j\omega_0 t})+Re(e^{j0}e^{j\omega_0 t})$$