Phase shifts of QPSK
In QPSK, the bit information is embedded in the phase of the carrier signal. For example, for bits 00, the phase shift is 45degree. This means the carrier signal $x(t) = cos(2\pi f_c t + \phi)  0 \lt t \lt T$, the phase shift $\phi = \pi / 4$. So $x(t) = cos(2\pi f_c t + \pi / 4) = cos(2\pi f_c t) cos(pi/4) - sin(2\pi f_ct)sin(pi/4)$. The Inphase component corresponding to $cos(2\pi f_c t)$ has factor of $cos(pi/4)$ which is what you see in the I component. Similar logic for Q component corresponding to $sin(2\pi f_c t)$. If the bits are 10, then I component will have factor of $cos(3 \pi/ 4)$ which will be $-\frac{1}{\sqrt{2}}$ so you will see phase shift of 180 degree due to the -1 factor.