Let $(B_t)$ a Brownian motion on $\mathbb R$. I want to show that $W_t=(W_t^1,W_t^2):=(\cos(B_t),\sin(B_t))$ is a Brownian motion on the circle $\mathbb S^1$. I have to prove that the generator in a coordinate system $(x_1,...,x_m)$ is $$\frac{1}{2}\Delta :=\frac{1}{2}\frac{1}{\sqrt{\det(g)}}\sum_{i=1}^m\partial _{x_i} \left(\sqrt{\det(g)}\sum_{j=1}^m g^{ij}\partial _{x_j}\right)$$ where $(g_{ij})^{-1}=(g^{ij})$.
Attempt: Let $s\in \mathbb S^1$. I can write it in the coordinate $(\theta)$ as $s=(\cos(\theta ),\sin(\theta ))$. Since $d\theta =-\sin(\theta )\partial _x+\cos(\theta )\partial _y$ I get that $d\theta ^2=1$ and thus $g=(1)$ (the matrix $1\times 1$ with the entry $1$). So, I have to prove that the generator is $\frac{1}{2}\frac{d^2}{d\theta ^2}$. Let $f\in \mathcal C^2$. First, $$dW_t^1=-\sin(B_t)dB_t-\frac{1}{2}\cos(B_t)dt$$ and $$dW_t^2=\cos(B_t)dB_t-\frac{1}{2}sin(B_t)dt.$$
Therefore, $$df(W_t)=\partial _x fdW_t^1+\frac{1}{2}(\partial _{xx}f+\partial _{yy}f)dt$$ $$=(-\sin(B_t)\partial _xf dB_t+\cos(B_t)\partial _yfdB_t+\frac{1}{2}\Bigg(-\cos(B_t)\partial _x f-\sin(B_t)\partial _y f+\partial _{xx}f+\partial _{yy}f\Bigg)dt$$
and thus the generator is $\Bigg(-\cos(B_t)\partial _x f-\sin(B_t)\partial _y f+\partial _{xx}f+\partial _{yy}f\Bigg)$ which look far away from $\frac{1}{2}\frac{d^2}{d\theta ^2}$.
What's wrong ?