Does "keying on" a sine wave at a zero-crossing reduce its bandwidth?
It is instructive to have a look at the expression for the spectrum of a truncated sine wave to see how the choice of the window length affects the decay of the spectrum. Let's define a windowed sinusoid $s(t)$:

$$s(t)=\sin(2\pi f_0t)\cdot\text{rect}(t,T)=\begin{cases}\sin(2\pi f_0t),&amp;-T/2&lt;t&lt;T/2\\0,&amp; \text{otherwise}\end{cases}\tag{1}$$

The Fourier transform of $s(t)$ is the convolution of the transforms of the sinusoid and the rectangular window:

$$\begin{align}S(f)&amp;=\frac{1}{2j}\left[\delta(f-f_0)-\delta(f+f_0)\right]\star T\text{sinc}(Tf)\\&amp;=\frac{T}{2j}\left[\text{sinc}(T(f-f_0))-\text{sinc}(T(f+f_0))\right]\tag{2}\end{align}$$

With $\text{sinc}(x)=\sin(\pi x)/(\pi x)$ we get from $(2)$

$$\begin{align}S(t)&amp;=\frac{T}{2 j}\left[\frac{\sin(\pi T(f-f_0))}{\pi T(f-f_0)}-\frac{\sin(\pi T(f+f_0))}{\pi T(f+f_0)}\right]\\&amp;=\frac{1}{2\pi j}\frac{(f+f_0)\sin(\pi T(f-f_0))-(f-f_0)\sin(\pi T(f+f_0))}{f^2-f_0^2}\\=&amp;\frac{1}{2\pi j(f^2-f_0^2)}\Big\{f\big[\sin(\pi T(f-f_0))-\sin(\pi T(f+f_0))\big]+f_0\big[\sin(\pi T(f-f_0))+\sin(\pi T(f+f_0))\big]\Big\}\tag{3}\end{align}$$

The spectrum $(3)$ decays as $\propto 1/f$ because we have $f^2$ in the denominator and $f$ in the first term of the numerator. If the window length $T$ is chosen such that an integer number of periods of the sinusoid are inside the window, then the limits of the window will be exactly at zero crossings of the sinusoid. This condition is equivalent to $Tf_0=n,\;n\in\mathbb{Z}$. With this condition the first term in the numerator of $(3)$ vanishes because the arguments of the two sinusoids differ by $2\pi n$, and, consequently, both sinusoids are identical. As a consequence, in that case the $f$ in the numerator disappears and the spectrum $S(f)$ will decay as $\propto 1/f^2$ instead of $\propto 1/f$.

In sum, if the windowing does not cause any discontinuities (i.e., the window limits occur at zero crossings of the sinusoid), the spectrum of the windowed sinusoid will decay as $\propto 1/f^2$. This is generally the case for continuous functions. Such functions can of course decay even faster, but only if they also have continuous derivatives, which is not the case for a windowed sinusoid. If the windowed sinusoid is discontinuous, i.e., the window limits do not occur at zero crossings, then the corresponding spectrum only decays as $\propto 1/f$.