# Dealing with a sequencing issue

1. Feb 18, 2014

### Slightly

When solving a Fourier Series, you can run into things that have the following structure.

cos(n*pi)

cos(n*2pi)

sin(n*pi/2)

These expressions can be rewritten in terms of powers of -1 or as well as seeing that some terms drop to zero or are always the same value.

cos(n*pi) = (-1)^n

cos(n*2pi) = 1

sin(n*pi/2) = (-1)^((n-1)/2)

My question comes down to: What do you get or how to figure out what happens with this:

sin(n*pi/3)

I know that the constant that always pops out is sqrt(3)/2, but the sign and pattern of this value starting at n=1 is as follows.

+ + 0 - - 0 + + 0 - - 0 ...etc

Is there a way to write this in powers of 1. I know that the 3n's are all equal to 0, but what about the other numbers?

Another way to avoid this is to not try to find a different way of writing sin(n*pi/3) since it is acceptable to just use it as part of the function.

But is there a way to write it?

2. Feb 19, 2014

### HallsofIvy

I presume you know that $sin(\pi/3)= \frac{\sqrt{3}}{2}$ and $cos(\pi/3)= \frac{1}{2}. From there you can use the sum rules: $sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)$ and $cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)$. For example, $sin(2\pi/3)= sin(\pi/3+ \pi/3)= sin(\pi/3)cos(\pi/3)+ cos(\pi/3)sin(\pi/3)= (\frac{\sqrt{3}}{3})(\frac{1}{2})+ (\frac{1}{2})(\frac{\sqrt{3}}{2}= \frac{\sqrt{3}}{2}$ and $cos(2\pi/3)= cos(\pi/3)cos(\pi/3)- sin(\pi/3)sin(\pi/3)= \frac{1}{4}- \frac{3}{4}= -\frac{1}{2}$. Of course, $sin(3\pi/3)= sin(\pi)= 0$ and cos(3\pi/3)= cos(\pi)= -1$.

$sin(4\pi/3)= sin(\pi+ \pi/3)= sin(\pi)cos(\pi/3)+ cos(\pi)sin(\pi/3)= 0(1/2)+ (-1)(\frac{\sqrt{3}}{2})= -\frac{\sqrt{3}}{2}$ and $$cos(4\pi/3)= cos(\pi+ \pi/3)= cos(\pi)cos(\pi/3)- sin(\pi)sin(\pi/3)= (-1)(1/2)- (0)(\sqrt{3}{2})= -1/2$$.

$sin(5\pi/3)= sin(\pi+ 2\pi/3)= sin(\pi)cos(2\pi/3)+ cos(\pi)sin(2\pi/3)= (0)(-1/2)+ (-1)(\frac{\sqrt{3}}{2})= -\frac{\sqrt{3}}{2}$ and cos(5\pi/3)= cos(\pi)cos(2\p/3)- sin(\pi)sin(2\pi/3)= (-1)(-1/2)- (0)(\frac{\sqrt{3}}{2})= 1/2[/itex].

Of course, $sin(6\pi/3)= sin(2\pi)= 0$ and $cos(6\pi/3)= cos(2\pi)= 1$

And now every thing repeats.