Sum of Sine Series

Date: 07/17/2008 at 23:15:35
From: Lauren
Subject: Sum of Sines and Cosines

I am completely lost with how to even get started on this question!

Show that for any integer n >= 1, the sum from k = 1 to k = n of
sin(kt) = [cos(1/2t) - cos(n + 1/2)t]/[2 sin(1/2t)]

The hint says to write sin(kt) sin(1/2t) as a linear combination of
cosine functions, but I am unsure how to even do that.  Is there a
formula for sin kt or sin 1/2t that involves cosines?  Will the half
angle formula for cosines help?  It involves a square root so that
doesn't seem right.  I do know the half angle formula.  Also, I know 
that there is a formula for sin 2t, but is there a general formula 
for sin(kt)?

Date: 07/18/2008 at 10:19:48
From: Doctor Ali
Subject: Re: Sum of Sines and Cosines

Hi Lauren!

Thanks for writing to Dr. Math.

So you want to find,

  Sin(t) + Sin(2t) + Sin(3t) + ... + Sin(nt)

Does that make sense?

The trick is to multiply and divide the series by 2 Sin(t/2).  That 
is,


  2 Sin(t/2) [ Sin(t) + Sin(2t) + Sin(3t) + ... + Sin(nt) ]
  ---------------------------------------------------------
                        2 Sin(t/2)

If you distribute that 2 Sin(t/2) to all terms in the numerator, 
you'll be able to write the expression in terms of cosines.  

For example, the first term will be:

  2 Sin(t/2) Sin(t)

We know that:

  2 Sin(a) Sin(b) = Cos(a-b) - Cos(a+b)

Are you familiar with the formula?

So, by assuming

  a = t/2
  b = t

we can write

  2 Sin(t/2) Sin(t) = Cos(t/2) - Cos(3t/2).

For the second term we have:

  2 Sin(t/2) Sin(2t) = Cos(3t/2) - Cos(5t/2)

If you continue like this, you'll have:

  2 Sin(t/2) Sin(t)  = Cos(t/2)  - Cos(3t/2)
  2 Sin(t/2) Sin(2t) = Cos(3t/2) - Cos(5t/2)
  2 Sin(t/2) Sin(3t) = Cos(5t/2) - Cos(7t/2)
  2 Sin(t/2) Sin(4t) = Cos(7t/2) - Cos(9t/2)
  .
  .
  .
  2 Sin(t/2) Sin(nt) = Cos((2n-1)t/2) - Cos((2n+1)t/2)

Adding all of the above equalities together, you'll see that the terms
on the right-hand sides will all cancel out, except Cos(t/2) and
-Cos((2n+1)t/2).

That is:
               n
              ---
  2 Sin(t/2)  )   Sin(kt)  =  Cos(t/2) - Cos((2n+1)t/2)
              ---
              k=1

Does that make sense?

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 07/18/2008 at 15:44:04
From: Lauren
Subject: Thank you (Sum of Sines and Cosines)

Thank you so much for your help!  This helps a lot!  I am a middle
school math teacher taking a graduate class and Ask Dr. Math is a huge
help.  Is there any way I can pay back by volunteering to answer some
lower level questions?