Fake Brownian motion – not Gaussian


Let $G$ be a standard normal random variable and define two standard Brownian motions $(W_t)_{t ge 0}$, $&$ $(B_t)_{t ge 0}$. Assume $G, (B_t)$ and $(W_t)$ are independent.

Moreover, define that process $Y_t$ by
$$
Y_t =
begin{cases}
B_t, & 0 le t le 1 \
sqrt{t}big(B_1 cos(W_{log t})+ G sin(W_{log t})big) & t ge 1
end{cases}
$$

Show that ${Y_t : t ge 0 }$ is not Brownian motion by proving that it is not Gaussian (this is called fake Brownian motion).

My attempt:

$Y_e – Y_1 = sqrt{e}(B_1cos(W_1)+Bsin(W_1)-1) = B_1(sqrt{e} cos(W_1) -1) + G sin(W_1)$. I know that any linear combination of independent normal random variables is also normal. However, $cos(a)$ and $sin(a)$ are not linear transformations. I’m not quite sure how to prove that this isn’t Gaussian because I don’t know the distribution of $cos(W_1)$ and $sin(W_1)$. Is there another way to show this?