# Explain the arithmetics of this trigonometrical expression

$$(\frac{d_1}{2} + \frac{d_1}{2} \cdot \cos(2x))+(\frac{d_2}{2} \cdot \sin(2x)) = \frac{d_1}{2}.$$

Can someone explain how the arithmetics works here? (My teacher has only noted that "filtration removes $\cos(2x)$ and $\sin(2x)$")

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So $d_1\cos(2x) + d_2 \sin(2x) = 0$? What are $d_1, d_2$? Just constants? – user2468 Aug 23 '12 at 14:50
If you are writing this equation then there must be some relationship b/w d1 and d2 – Rahul Taneja Aug 23 '12 at 15:23
Are you looking to solve for $d_1$ and $d_2$? Because, in particular, for $x=0$, the equation is false unless $d_1=0$, in which case $d_2=0$ can be seen by taking $x=\pi/4$, for instance. – Michael Boratko Aug 23 '12 at 15:31
Jennifer - Yes $d_1$ and $d_2$ are just constants.   Rahul - The relationship between d1 and d2 are that they are 2 different carriers transmitted over the same channel and through the same signal (see my dropbox images for further mathematical explaination.   Michael - No, I'm not looking to solve $d_1$ and $d_2$, I just want to know how you simplify the LHS expression into the simplified RHS expression, what is the arithmetics behind it? – user38541 Aug 23 '12 at 16:05
Images:  modulation - dl.dropbox.com/u/16952797/temp_stuff/KT/QAM/20120823_174940.jpg demodulation - dl.dropbox.com/u/16952797/temp_stuff/KT/QAM/20120823_175218.jpg – user38541 Aug 23 '12 at 16:08

## 1 Answer

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.

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Alright, that makes sense. Thank you! – user38541 Aug 23 '12 at 17:23
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