$(\frac{d_1}{2} + \frac{d_1}{2} \cdot \cos(2x))+(\frac{d_2}{2} \cdot \sin(2x)) = \frac{d_1}{2}.$
Can someone explain how the arithmetics works here? (My teacher has only noted that "filtration removes $\cos(2x)$ and $\sin(2x)$")
$(\frac{d_1}{2} + \frac{d_1}{2} \cdot \cos(2x))+(\frac{d_2}{2} \cdot \sin(2x)) = \frac{d_1}{2}.$
Can someone explain how the arithmetics works here? (My teacher has only noted that "filtration removes $\cos(2x)$ and $\sin(2x)$")
As has been pointed out in comments, for this equation to hold for all $x$, we need to have $d_1=d_2=0$. The reference to being removed by filtration doesn't refer to an arithmetic operation inherent in this equation, but to the application of a low-pass filter indicated in the second image linked to in a comment.