Fourier Transform Form: two sin components & a phase shift & a magnitude for only one term
Fourier Transform is a linear one, so you can make use of superposition principle:

$$ \mathscr{F} [\alpha x(t) + \beta y(t)] = \alpha \mathscr{F}[x(t)] + \beta \mathscr{F}[y(t)] $$

So for the first component $$x(t) = \sin \left( 2\pi \cdot 1000 \cdot t\right)$$

by definition:

$$\mathscr{F}\left[\sin(2\pi f_0 t + \phi) \right] = \dfrac{i}{2} \left[ e^{-i \phi}\delta(f+f_0) - e^{i \phi}\delta(f-f_0) \right] $$

you get:

$$ \mathscr{F}[x(t)]=\dfrac{i}{2} \left[ \delta(f+1000) - \delta(f-1000) \right]  $$

Second component is a sinusoid with shifted phase, so the complex exponent represents that:

$$y(t) = \dfrac{1}{2} \sin \left( 2\pi \cdot 2000 \cdot t + \dfrac{3\pi}{4} \right)$$

has following Fourier Transform:

$$\mathscr{F}[y(t)] = \dfrac{1}{2}\dfrac{i}{2} \left[ e^{\dfrac{-3\pi i}{4}}\delta(f+2000) - e^{\dfrac{3\pi i}{4}}\delta(f-2000) \right] $$

By summing both results you get the Fourier Transform of your signal.