The Fourier transform of a periodic signal is zero everywhere except at integer multiples of the fundamental frequency. Using the Dirac delta impulse, it is straightforward to derive it from the Fourier series. Let $x(t)$ be a $T$-periodic function with Fourier coefficients $c_k$:
$$x(t)=\sum_{k=-\infty}^{\infty}c_ke^{jk\omega_0 t},\quad \omega_0=\frac{2\pi}{T}\tag{1}$$
With
$$\mathcal{F}\big\{e^{j\omega_0t}\big\}=2\pi\delta(\omega-\omega_0)\tag{2}$$
we get from $(1)$ and $(2)$
$$X(j\omega)=\mathcal{F}\big\{x(t)\big\}=2\pi\sum_{k=-\infty}^{\infty}c_k\delta(\omega-k\omega_0)\tag{3}$$
Reacting to a comment about the Laplace transform, it's important to realize that the Laplace transform of a truly periodic function doesn't exist. E.g., there is no Laplace transform of $\sin(\omega_0t)$ or $\cos(\omega_0t)$. If a Laplace transform table says otherwise, then what they really mean is the Laplace transform of $\sin(\omega_0t)u(t)$ or $\cos(\omega_0t)u(t)$, where $u(t)$ is the unit step function. These functions are not periodic because they are zero for $t<0$.
The Laplace transform of such a pseudo-periodic function satisfying $f(t+T)=f(t)$ for $T>0$ and $t>0$ is given by
$$F(s)=\frac{F_0(s)}{1-e^{-sT}}\tag{4}$$
where $F_0(s)$ is the Laplace transform of one period of $f(t)$ in the interval $[0,T]$:
$$F_0(s)=\int_0^Tf(t)e^{-st}dt\tag{5}$$
