Summation of sines and cosines questions

[mjordan2nd]

This is not a homework question per say, but rather a question I have about a text I am reading. In the text, they have defined

$$\sum E_{n} Cos \left( \frac{2 \pi X_{n}}{\lambda} \right) = E Cos \phi$$

If this is the case, is it fair to say that

$$\sum E_{n} Sin \left( \frac{2 \pi X_{n}}{\lambda} \right) = E Sin \phi$$

My text claims it is, though I can not figure out why. Any help would be appreciated.

 

[jbunniii]

Yes, it is fair to say, and the reason is Euler's identity,

$$e^{ix} = \cos(x) + i \sin (x)$$

Try starting with

$$\sum E_n \exp\left(\frac{i2 \pi X_n}{\lambda}\right)$$

and see what you come up with.

 

[mjordan2nd]

Ahh. Let me see if I can make an argument out of this.

$$\sum E_n \exp \left( \frac{i2 \pi X_n}{\lambda} \right) = \sum E_{n} Cos \left( \frac{2 \pi X_{n}}{\lambda} \right) + E_{n} i Sin \left( \frac{2 \pi X_{n}}{\lambda} \right)$$

But we can also say

$$\sum E_n \exp\left(\frac{i2 \pi X_n}{\lambda}\right) = Ee^{i \phi} = E cos \phi + i E sin \phi$$

By setting the real parts equal to each other and the imaginary parts equal to each other, we see that the definitions are consistent.

Right?

 

[jbunniii]

Looks good to me.

 

[micromass]

There have to be some contraints to this, as it doesn't hold in general. For example:

$$\cos(x)=\cos(-x)$$

but not

$$\sin(x)=\sin(-x)$$

 

[jbunniii]

Maybe I read more into the question than was actually there.

It is true that there is a solution $E, \phi$ that satisfies the two equations, and with the convention that E > 0, the solution is unique. (E = 0 is a degenerate exception, in which case $\phi$ can be anything.)

Reading the question more literally, it does not make sense to define $E, \phi$ by just the one equation

$$\sum E_{n} Cos \left( \frac{2 \pi X_{n}}{\lambda} \right) = E Cos \phi$$

as there are infinitely many combinations of $E, \phi$ that work.

 

[mjordan2nd]

Thanks for all the help, folks. Jbunniii your first interpretation was of the question was the one I intended.