Find the output $y(t)$ given the input and system response using Fourier Transform

I need to find the output $$y(t)$$ given the input $$x(t)=\cos(t)$$, and impulse response $$h(t)=u(t)$$ using the Fourier Transform.

I know that convolution in time domain is multiplication in the frequency domain, but since $$\cos(t)$$ has a Fourier Transform which involves the delta function, I am unable to figure out how to perform the multiplication in the frequency domain.

The answer in the time domain is $$y(t)=\sin(t)$$. I don't know how to get this. If someone could show the working, would be highly appreciated!

From an engineering perspective, if you assume that

$$\delta(s-a) \cdot 0 = 0$$

and thus

$$\delta(s-a)\cdot\delta(s-b) = 0 \quad a \ne b$$

then

\begin{align*}y(t) &= h(t) \ast x(t)\\ \\ &= u(t)\ast \cos(t)\\ \\ &= \mathscr{F}^{-1}\left\{\mathscr{F}\left\{u(t)\right\}\cdot\mathscr{F}\left\{\cos(t)\right\}\right\}\\ \\ &= \mathscr{F}^{-1}\left\{\dfrac{1}{2}\left[\dfrac{1}{i\pi s}+\delta(s)\right]\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right]\right\}\\ \\ &= \mathscr{F}^{-1}\left\{\dfrac{1}{2\pi i s}\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right]\right\}\\ \\ &= \int_{-\infty}^\infty \dfrac{e^{2\pi i st}}{2\pi i s}\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right] ds\\ \\ &= \dfrac{1}{2}\left[\dfrac{e^{2\pi i \frac{-1}{2\pi}t}}{2 \pi i \frac{-1}{2\pi}}+ \dfrac{e^{2\pi i \frac{1}{2\pi}t}}{2 \pi i \frac{1}{2\pi}}\right] \\ \\ &= \dfrac{e^{it}-e^{-it}}{2i}\\ \\ y(t) &= \sin(t) \\ \end{align*}

Technically, instead of using a hand-waving $$\delta(s-a)\delta(s-b) = 0$$ for $$a\ne b$$, I suppose I should have also done the two integrals with products of delta functions:

$$\int_{-\infty}^{\infty} \dfrac{e^{2\pi i st}}{4} \delta(s)\delta\left(s+\dfrac{1}{2\pi}\right)ds$$

$$\int_{-\infty}^{\infty} \dfrac{e^{2\pi i st}}{4} \delta(s)\delta\left(s-\dfrac{1}{2\pi}\right)ds$$

However, using the properties of the delta function, these will come out to be 0.