I never really understood why the amplitude of the Fourier transform does not represent the amplitude of a sin/cos signal of a specific frequency? The way the Fourier transform is typically defined, makes taking the Fourier transform of the sine function to be infinity at its frequency (delta function). Since the point of the Fourier transform is to find the coefficients (amplitudes) of the individual sine and cosine waves, why not define it by dividing the dot product of the function with the corresponding dot product of the basis functions? This way one gets the amplitude of the individual sine/cosine waves?
Example: $$ \begin{align} \left|\mathscr{F}(f(t))\right| &= \left|\frac{\int_{-\infty}^\infty f(t)\cos(\omega t)\,dx}{\int_{-\infty}^\infty \cos^2(\omega t)\,dx} + i\,\frac{\int_{-\infty}^\infty f(t)\sin(\omega t)\,dx}{\int_{-\infty}^\infty \sin^2(\omega t)\,dx}\right| \\[6pt] \left|\mathscr{F}(\sin(at))\right| &= \left|\frac{\int_{-\infty}^\infty \sin(at)\cos(\omega t)\,dx}{\int_{-\infty}^\infty \cos^2(\omega t)\,dx} + i\,\frac{\int_{-\infty}^\infty \sin(at)\sin(\omega t)\,dx}{\int_{-\infty}^\infty \sin^2(\omega t)\,dx}\right| \\[4pt] &= \left|\,0 + i\,\frac{\int_{-\infty}^\infty \sin(at)\sin(\omega t)\,dx}{\int_{-\infty}^\infty \sin^2(\omega t)\,dx }\right| \\[4pt] &= \begin{cases} 1, & \omega = a \\ 0, & \omega \neq a \end{cases} \end{align} $$
Instead of the usual Dirac delta functions which are infinity at $\omega = a$?
The Fourier transform
$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\int\limits_{-\infty}^\infty f(x)\, e^{-2 \pi i \omega x}\, dx\tag{1}$$
of any periodic function $f(x)$ doesn't converge in the usual sense (except when $f(x)=0$) because periodic functions don't decay as $|x|\to\infty$. Consequently the Fourier transform of a periodic function can only converge in a distributional sense.
If $f(x)$ is a periodic function it can be represented by the exponential Fourier series
$$f(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=-N}^N F_P(n)\, e^{i \frac{2 \pi}{P} n x}\right),\quad x\in\mathbb{R}\tag{2}$$
where $P$ is the period and
$$F_P(\omega)=\frac{1}{P} \int\limits_{-\frac{P}{2}}^{\frac{P}{2}} f(x)\, e^{-i \frac{2 \pi}{P} \omega x} \, dx\tag{3}$$
is the truncated Fourier transform of $f(x)$ (i.e. the integration limits are truncated).
If $f(x)$ is a non-periodic function it can be represented over the interval $a<x<a+P$ by the exponential Fourier series
$$f(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=-N}^N F_{a,P}(n)\, e^{i \frac{2 \pi}{P} n x}\right),\quad a<x<a+P\tag{4}$$
where $P$ is the period and
$$F_{a,P}(\omega)=\frac{1}{P} \int\limits_a^{a+P} f(x)\, e^{-i \frac{2 \pi}{P} \omega x} \, dx\tag{5}$$
is the truncated Fourier transform of $f(x)$.
As a trivial example, the function $f(x)=\cos(2 \pi x)$ can be represented as
$$f(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=-N}^N F_1(n)\, e^{i 2 \pi n x}\right),\quad x\in\mathbb{R}\tag{6}$$
where $P=1$ is the period and
$$F_1(\omega)=\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} f(x)\, e^{-i 2 \pi \omega x} \, dx=\frac{\omega \sin(\pi \omega)}{\pi \left(1-\omega^2\right)}\tag{7}$$
is the truncated Fourier transform of $f(x)=\cos(2 \pi x)$.
Note that $F_1(\omega)$ defined in formula (7) above evaluates to $0$ at integer values of $\omega$ except at $\omega=\pm 1$ where it has removable singularities since
$$\underset{\omega\to 1}{\text{lim}}\left(\frac{\omega \sin (\pi \omega)}{\pi \left(1-\omega^2\right)}\right)=\underset{\omega\to -1}{\text{lim}}\left(\frac{\omega \sin(\pi \omega)}{\pi \left(1-\omega^2\right)}\right)=\frac{1}{2}\tag{8}$$
and so the Fourier series for $f(x)=\cos(2 \pi x)$ can be expressed as
$$f(x)=\frac{1}{2} e^{-2 \pi i x}+\frac{1}{2} e^{2 \pi i x}=\cos(2 \pi x)\tag{9}.$$
In conclusion Fourier series are based on truncated Fourier transforms which evaluate to finite values at the fundamental frequency and its harmonics.