# Complex numbers equation/equality

1. Jan 27, 2014

### mahler1

The problem statement, all variables and given/known data.

Prove that, given constants $A_1,A_2, \phi_1$ and $\phi_2$, there are constants $A$ and $\phi$ such that the following equality is satisfied:

$A_1\cos(kx+\phi_1)+A_2\cos(kx+\phi_2)=A\cos(kx+\phi)$

The attempt at a solution.

I've tried to use the identity $\cos(u+v)=\cos(u)\cos(v)-\sin(u)\sin(v)$, so

$A_1\cos(kx+\phi_1)+A_2\cos(kx+\phi_2)=A_1[\cos(kx)\cos(\phi_1)-\sin(kx)\sin(\phi_1)]+A_2[\cos(kx)\cos(\phi_2)-\sin(kx)\sin(\phi_2)]$, and then, the expresion on the right equals to

$[A_1\cos(\phi_1)+A_2\cos(\phi_2)]\cos(kx)-[A_1\sin(\phi_1)+A_2\sin(\phi_2)]\sin(kx)$.

I am trying to write this last expression as something of the form $A\cos(kx+\phi)$ but I don't know how to, I would appreciate any suggestion to arrive to the desired expression.

2. Jan 27, 2014

### Dick

Here http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Linear_combinations is the sort of identity you need expressed for real functions. Can you figure out how to derive a similar sort of thing for complex functions? You don't need to actually find the complex magnitude or the phase, you just have to show they exist. I'm actually not sure I can now, it's getting kind of late. So I'm counting on you.

Last edited: Jan 27, 2014
3. Jan 28, 2014

### Dick

Ok, here's a little more. You already have it in the form of a linear combination of $sin(kx)$ and $cos(kx)$. Using the definitions of complex sin and cos you could equally well write that as $C_1 e^{ikz}+C_2 e^{-ikz}$. Now let's introduce a complex phase $\phi$, by computing it such that $C_1 e^{-i \phi}=C_2 e^{i \phi}$. Can you see how it goes from here?

4. Jan 28, 2014

### mahler1

As you've said, I could express $[A_1\cos(\phi_1)+A_2\cos(\phi_2)]\cos(kx)-[A_1\sin(\phi_1)+A_2\sin(\phi_2)]\sin(kx)$ as $C1e^{ikx}+C_2e^{-ikx}$, if I call $k_1=A_1\cos(\phi_1)+A_2\cos(\phi_2)$ and $k_2=A_1\sin(\phi_1)+A_2\sin(\phi_2)$, then $C_1,C_2$ must satisfy $C_1=\dfrac{k_1-ik_2}{2}$ and $C_2=\dfrac{k_1+ik_2}{2}$.

I didn't get your last suggestion, suppose I find $\phi$ such that $C_1 e^{-i \phi}=C_2 e^{i \phi}$, and how would this help me to write the original expression in the form $A\cos(kx+\phi)$? Sorry if it is too obvious.

5. Jan 28, 2014

### AlephZero

If you write $A_1 \cos(kx + \phi_1) = A_1\cos\phi_1\cos kx - A_1\sin\phi_1\sin kx = P_1\cos kx + P_2\sin kx$ say, and similarly for the other terms, you have to show that the result is true for any value of $\cos kx$ and $\sin kx$. You can do that by equating the coefficients of $\cos kx$ and $\sin kx$.

Or, if you use $e^{ikx} = \cos kx + i\sin kx$, this is the same as equating both the real and imaginary parts of $e^{ikx}$.

6. Jan 28, 2014

### Dick

Ok, I'll tell you what the strategy is. Given I can write the expression as $C_1e^{ikx}+C_2e^{-ikx}$. I would very much like to find a phase $\phi$ such that $C_1e^{ikx}+C_2e^{-ikx}=De^{i(kx+\phi)}+De^{-i(kx+\phi)}$, what equation would I need to solve for $\phi$? That actually probably would have been a better clue.

Last edited: Jan 28, 2014
7. Jan 28, 2014

### Dick

All of the coefficients here are complex. So it's not quite the same as equating real and imaginary parts. It's equating coefficients of the linearly independent functions $e^{ikx}$ and $e^{-ikx}$.