Phasors: why is my book using sines?
I also saw the first definition in my education:

$$v(t) = \mathcal{R}\left\{\underline{V}\cdot e^{j\omega t}\right\} = \mathcal{R}\left\{(V\cdot e^{j\phi})\cdot e^{j\omega t}\right\} = V\cdot\cos(\omega t+\phi)$$

But as long as you are consistent with your formulation, both will lead to the same results when analyzing problems. The phasor amplitude will now represent the RMS value of the sinusoidal signal, and the phase will be relative to a sine wave (90 degrees shifted compared to the cosine).

[EDIT] If you're also interested in how exactly they're linked:

Both definitions are still strongly linked. Using the second definition is basically identical to

$$\begin{align}
v(t) &amp;= \mathcal{I}\left\{\underline{V}'\cdot \sqrt{2} e^{j\omega t}\right\}\\
&amp;= \mathcal{I}\left\{(V'\cdot e^{j\phi'})\cdot \sqrt{2} e^{j\omega t}\right\}\\ &amp;=\sqrt{2}V'\cdot \sin(\omega t + \phi')\\
&amp;= (\sqrt{2}V')\cdot \cos\left(\omega t + \left(\phi' - \frac{\pi}{2}\right)\right)
\end{align}$$

And you can find that

$$\begin{align}
V &amp;= \sqrt{2}V'\\
\phi &amp;= \phi' - \frac{\pi}{2}\\
\underline{V} &amp;= Ve^{j\phi} = (\sqrt{2}V')e^{j(\phi'-\frac{\pi}{2})} = -j\sqrt{2}\cdot \underline{V'}
\end{align}$$