Simplification of a complex exponential

[fonz]

Homework Statement

Is there a way to simplify the following expression?

$[cos(\frac {n \pi} 2) - j sin(\frac {n \pi} 2)] + [cos(\frac {3n \pi} 2) - j sin(\frac {3n \pi} 2)]$

Homework Equations

$e^{jx} = cos(x) + j sin(x)$

The Attempt at a Solution

$cos(\frac {n \pi} 2)$ and $cos(\frac {3n \pi} 2)$ produce the same result for all values of $n$

Likewise, $- j sin(\frac {n \pi} 2)$ and $- j sin(\frac {3n \pi} 2)$ produce the same result for all values of $n$

 

[fresh_42]

So if you know this (sure the sine formula is correct?), you can substitute the three folds of the angles by the simple angles. Next you wrote $e^{jx} = \cos(x)+j\sin(x)$. Now what is $e^{-jx}$ and how are $e^{jx}$ and $e^{-jx}$ related? All these give you some simplifications. Whether they are those you're looking for depends on what you mean by simplification.

 

[Ray Vickson]

Homework Statement

Is there a way to simplify the following expression?

$[cos(\frac {n \pi} 2) - j sin(\frac {n \pi} 2)] + [cos(\frac {3n \pi} 2) - j sin(\frac {3n \pi} 2)]$

Homework Equations

$e^{jx} = cos(x) + j sin(x)$

The Attempt at a Solution

$cos(\frac {n \pi} 2)$ and $cos(\frac {3n \pi} 2)$ produce the same result for all values of $n$

Likewise, $- j sin(\frac {n \pi} 2)$ and $- j sin(\frac {3n \pi} 2)$ produce the same result for all values of $n$

Is $n$ supposed to be an integer?

BTW: don't write $cos(x)$; write $\cos(x)$ instead. The first is ugly and hard to read, the second is pleasing and easy to read. To achieve this, just put a "\" in front of "sin", "cos", and most other standard functions---that is the way LaTeX was designed to be used. In other words, instead of using "sin", "cos", "tan", "log", "lim", etc., etc., use "\sin", "\cos', "\tan", "\og", "\lim", etc.

 

[fonz]

Sorry I forgot to mention, $n$ is an integer. I am wonder if there is any way to simplify in this case?

Is there anything wrong with the following?

$[\cos(\frac {n \pi} 2) - j \sin(\frac {n \pi} 2)] + [\cos(\frac {3 n \pi} 2) - j \sin(\frac {3 n \pi} 2)] = 2[\cos(\frac {n \pi} 2) - j \sin(\frac {n \pi} 2)]$

 

[fresh_42]

What is $\sin (\varphi + \pi)$ and what if applied to $\sin (\dfrac{3n\pi}{2})=\sin(\dfrac{n\pi}{2}+\pi)$?

 

[fonz]

What is $\sin (\varphi + \pi)$ and what if applied to $\sin (\dfrac{3n\pi}{2})=\sin(\dfrac{n\pi}{2}+\pi)$?

$\sin(\varphi + \pi) = \sin(\varphi)$

$\sin(\frac {3n\pi} 2) = \sin(\frac {n\pi} 2)$

In which case my answer above should be correct.

 

[fresh_42]

$\sin(\varphi + \pi) = \sin(\varphi)$

$\sin(\frac {3n\pi} 2) = \sin(\frac {n\pi} 2)$

In which case my answer above should be correct.

Yes, in this case the answer would be correct. However, $\sin (\varphi + \pi) = - \sin (\varphi)$, so you missed a minus sign. The result will still have a $\cos (\dfrac{n\pi}{2})$ term, which can be expressed as function of $\cos (n\pi)$ if you like.

 

[fonz]

Yes, in this case the answer would be correct. However, $\sin (\varphi + \pi) = - \sin (\varphi)$, so you missed a minus sign. The result will still have a $\cos (\dfrac{n\pi}{2})$ term, which can be expressed as function of $\cos (n\pi)$ if you like.

Very helpful thank you.

 

[SammyS]

Is there a way to simplify the following expression?

$\displaystyle \left [\cos\left(\frac {n \pi} 2\right) - j \sin\left(\frac {n \pi} 2\right) \right ] + \left[\cos\left(\frac {3n \pi} 2\right) - j \sin\left(\frac {3n \pi} 2\right) \right]$

Homework Equations

$\displaystyle e^{jx} = \cos(x) + j \sin(x)$

I suggest using the exponent form to simplify this expression.

 

[FactChecker]

For particular values of integers n, you can put specific constants in for the sin and cos. So you can give a few cases that give the same constant value for each case.