MATLAB: Does FFT subtract PI/2 phase shift for sine wave

fftphase shiftsine

I know that very similar questions have already been asked, but I'm still confused. If I do the following:
t = 0:0.0001:1.0;
signal = 10*sin(2*pi*10*t-pi/2);
n = length(signal); 
z = fft(signal, n); %do the actual work
%generate the vector of frequencies
halfn = n / 2;
deltaf = 1 / ( n / ScanRate);
frq = (0:(halfn-1)) * deltaf;
% convert from 2 sided spectrum to 1 sided
%(assuming that the input is a real signal)
amp(1) = abs(z(1)) ./ (n);
amp(2:halfn) = abs(z(2:halfn)) ./ (n / 2);
phase = angle(z(1:halfn));
When I plot the phase versus the frequency, I get a phase shift of about -pi at a frequency of 10 Hz. I believe that this is coming from the fact that the sine wave is shifted pi/2 from the cosine wave. However, if this was an arbitrary signal, I wouldn't know if it was sine or cosine input. And, if I use this phase shift directly with a sine wave to do the inverse transform, it won't match my original input signal. Does this mean that the basis function in some sense is a cosine wave?
Will someone please clear this confusion up? thanks so much.

Best Answer

Think of it this way. Cosine is the real-part of the complex exponential, whereas sine is the real-part of the complex exponential multiplied by -i. And -i is equivalent to exp(-i*pi/2). So that means that sine is out-of-phase by -pi/2 compared with cosine. In other words,
   sin(w*t) = cos(w*t - pi/2)
So that means the phase of cosine is 0, whereas the phase of sine is -pi/2.
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