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Selling Papayas

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I apologize in advance if any formatting is weird; this is my first time posting. If I am breaking any rules with the formatting or if I am not providing enough detail or if I am in the wrong sub-forum, please let me know.

Using Euler's formula : e

##\displaystyle \sin(x)=\frac {e^{jx} - e^{-jx}} {2j}##

##\displaystyle \cos(x)=\frac {e^{jx} + e^{-jx}} {2}##

write the following in terms of sin & cos:

I posted the ones given in the problem in part 1, and the only other one I used is ##j = e^{j \frac {π} {2}}##

This is probably obvious, but my classes use j instead of i to represent the imaginary number.

So for part 1, I think I solved it counter-intuitively (Wolfram Alpha says my solution is equivalent but its extremely long and I still feel as if I'm missing some key conversion)

I converted the j's into ##e^j\frac{π}{2}##, multiplied the exponentials together (adding the exponents in each term) and ended up with the following:

For part 2, I am getting stumped pretty quickly. I've tried to apply the methods I used in part 1 to solve this, but can't seem to pull out the k and the π from the denominator.

What I've done so far:

1. Homework Statement1. Homework Statement

Using Euler's formula : e

^{jx}= cos(x) + jsin(x) and the exponential representations of sin & cos, which are derived from Euler's formula:##\displaystyle \sin(x)=\frac {e^{jx} - e^{-jx}} {2j}##

##\displaystyle \cos(x)=\frac {e^{jx} + e^{-jx}} {2}##

- ##\displaystyle je^{j \frac {π} {4}t} - je^{-j \frac {π} {4}t} + 2e^{j \frac {5π} {4}t} + 2e^{-j \frac {5π} {4}t}##
- ##\displaystyle \frac {3} {4jkπ} (1-e^{-jkπ})^2##

## Homework Equations

I posted the ones given in the problem in part 1, and the only other one I used is ##j = e^{j \frac {π} {2}}##

This is probably obvious, but my classes use j instead of i to represent the imaginary number.

## The Attempt at a Solution

So for part 1, I think I solved it counter-intuitively (Wolfram Alpha says my solution is equivalent but its extremely long and I still feel as if I'm missing some key conversion)

I converted the j's into ##e^j\frac{π}{2}##, multiplied the exponentials together (adding the exponents in each term) and ended up with the following:

##[\cos(\frac {π} {2} + \frac {π} {4}t) + j\sin(\frac {π} {2} + \frac {π} {4}t)] - [\cos(\frac {π} {2} - \frac {π} {4}t) + j\sin(\frac {π} {2} - \frac {π} {4}t)] + 2[\cos(\frac {5π} {4}t) + j\sin(\frac {5π} {4}t)] + 2[\cos(\frac {-5π} {4}t) + j\sin(\frac {-5π} {4}t)]##

As you can see, it's a lot, but I am confident I at least have *a*right solution.For part 2, I am getting stumped pretty quickly. I've tried to apply the methods I used in part 1 to solve this, but can't seem to pull out the k and the π from the denominator.

What I've done so far:

##\displaystyle \frac {3} {4jkπ} (1-e^{-jkπ})^2##

##\displaystyle= \frac {3} {4jkπ} (1-e^{-jkπ})(1-e^{-jkπ})##

##\displaystyle= \frac {3} {4jkπ} (1 - 2e^{-jkπ} + e^{-2jkπ})##

##\displaystyle= \frac {3} {4jkπ} - \frac {3e^{-jkπ}} {2jkπ} + \frac{3e^{-2jkπ}} {4jkπ}##

##\displaystyle= \frac {3} {4jkπ} (1-e^{-jkπ})(1-e^{-jkπ})##

##\displaystyle= \frac {3} {4jkπ} (1 - 2e^{-jkπ} + e^{-2jkπ})##

##\displaystyle= \frac {3} {4jkπ} - \frac {3e^{-jkπ}} {2jkπ} + \frac{3e^{-2jkπ}} {4jkπ}##

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