Line spectrum of a given modulated signal

Given: $s(t)=50\cos(2\pi10^6t+60°)-15\cos(2\pi10^6t-60°)+20\sin(4\pi10^4t-190°)+\sin(2\pi10^4t+190°)$

To draw line spectra, write every term as a cosine with positive amplitude, and combine terms of the same frequency.

Step 1: 1 MHz terms (phasor addition)

$$\begin{aligned} 50\angle60°&=25+j43.30\\ -15\angle-60°&=-7.5+j12.99\\ \text{Sum}&=17.5+j56.29=58.95\angle72.73° \end{aligned}$$

So $50\cos(\cdot+60°)-15\cos(\cdot-60°)=58.95\cos(2\pi10^6t+72.73°)$.

Step 2: 20 kHz term ($4\pi10^4t=2\pi\cdot2\times10^4t$)

Using $\sin x=\cos(x-90°)$:

$$20\sin(\omega t-190°)=20\cos(\omega t-280°)=20\cos(2\pi\,20000\,t+80°)$$

Step 3: 10 kHz term

$$\sin(\omega t+190°)=\cos(\omega t+100°)=\cos(2\pi\,10000\,t+100°)$$

Result

$$s(t)=\cos(2\pi10^4t+100°)+20\cos(2\pi\,2{\times}10^4t+80°)+58.95\cos(2\pi10^6t+72.73°)$$

For the two-sided spectrum, each line splits into two of half amplitude at $\pm f$ (0.5, 10, 29.47) with phase $+\theta$ at $+f$ and $-\theta$ at $-f$.

 One-sided amplitude spectrum
 58.95 |                         |
       |                         |
    20 |     |                   |
     1 |  |  |                   |
       +--+--+------//-----------+--> f
         10  20 kHz            1 MHz

 One-sided phase spectrum (deg)
   100 |  o
    80 |     o
 72.73 |                         o
       +--+--+------//-----------+--> f
         10  20 kHz            1 MHz