I'm not totally sure by what you mean. If you mean get rid of the sine wave meaning just don't plot it:
fs = 44100; % Sampling frequency
Ts = 1/fs; % Time step.
t2 = -3 * pi : Ts : 3 * pi; % -3pi - 3pi s with time step Ts
% Original Sawtooth Waveform
x2 = (t2 + 2 * pi).*(t2 < -pi) + t2.*((-pi <= t2) & (t2 <= pi)) + (t2 - 2 * pi).*(t2 > pi);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
figure(2);
plot(t2, x2);
ylim([-4 4])
Otherwise, it looks like the sine wave you're talking about is using a Fourier series to construct a sawtooth wave using additive synthesis (ref). I made some slight modifications to this part of your script to follow the equation I linked in my reference. Synthesizing the sawtooth wave 8 times (k=8), we get:
fs = 44100; % Sampling frequency
Ts = 1/fs; % Time step.
t2 = -3 * pi : Ts : 3 * pi; % -3pi - 3pi s with time step Ts
% Original Sawtooth Waveform
x2 = (t2 + 2 * pi).*(t2 < -pi) + t2.*((-pi <= t2) & (t2 <= pi)) + (t2 - 2 * pi).*(t2 > pi);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
T0 = 2*pi;
f0 = 1/T0;
y2 = 0;
a = 2*pi;
for k = 1: 8
y2 = y2 + ((-1)^k)*sin(2*pi*k*f0*t2)/k;
end
y2 = a*(0.5-(1/pi)*y2);
figure();
plot(t2, x2, t2, y2);
Let's try k = 80, see that the more iterations you use the closer you get to the original sawtooth waveform:
fs = 44100; % Sampling frequency
Ts = 1/fs; % Time step.
t2 = -3 * pi : Ts : 3 * pi; % -3pi - 3pi s with time step Ts
% Original Sawtooth Waveform
x2 = (t2 + 2 * pi).*(t2 < -pi) + t2.*((-pi <= t2) & (t2 <= pi)) + (t2 - 2 * pi).*(t2 > pi);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
T0 = 2*pi;
f0 = 1/T0;
y2 = 0;
a = 2*pi;
for k = 1: 80
y2 = y2 + ((-1)^k)*sin(2*pi*k*f0*t2)/k;
end
y2 = a*(0.5-(1/pi)*y2);
figure();
plot(t2, x2, t2, y2);
For some reason my amplitude is shifted by pi. Not totally sure why but the shape itself seems correct.
