Expert: Paul Klarreich - 11/26/2005

Question
Hi,

I was wondering how I can prove that

The the summation of
sinX+sin2X+sin3X...sinnX to n terms
=sin(((n+1)/2)x)*sin((n/2)x)/sinx/2

Thanks.

Answer
Hi, KJ,

This will involve the following:

1. Mathematical induction.  (If you don't know what this is, send me another message and I'll explain.)

2. The following 'standard' trigonometric formulas.  (Standard, he says -- har, de har har har.)  If you want to look them up in your trig book, use the associated names.  Of course, there are lots of these.

                       A + B       A - B
cos A - cos B = -2 sin (-----) sin (-----)    
                         2           2
That is a SUM-TO-PRODUCT formula.

             cos(A - B) - cos(A - B)
sin A sin B = -----------------------  [PRODUCT-TO-SUM]
                       2

sin(-A) = - sin A    [SINE IS AN ODD FUNCTION]

-----------------------------------------------------
Ready?  Here we go.  To prove the theorem by M.I, you first verify the theorem for n = 1, which I leave to you.

Then we ASSUME it is true for  n = k, and try to use that to prove it for n = k+1.  We write those things out:

                                    k+1        k
                                sin(---x) sin(---x)
                                     2         2
sin x + sin 2x + ... + sin(kx) = --------------------
                                      sin(x/2)

THAT IS ASSUMED TO BE TRUE.


n=k+1:                                  k+2       k+1
                                   sin(---x) sin(---x)
                                        2         2
sin x + sin 2x + ... + sin((k+1)x) = --------------------
                                         sin(x/2)

THAT IS TO BE PROVED.

To prove it, we note that the left side (summation) is just the n=k sum plus an extra term.  So we take the left side and put in the thing assumed to be true, and we get:

sin x + sin 2x + ... + sin((k+1)x) =
sin x + sin 2x + ... + sin(kx) + sin((k+1)x) =

   k+1        k
sin(---x) sin(---x)
    2         2           
-------------------- + sin((k+1)x)  = with LCD= sin(x/2)
     sin(x/2)


   k+1        k
sin(---x) sin(---x) + sin((k+1)x) sin(x/2)
    2         2           
------------------------------------------
             sin(x/2)


Now we start using those formulas.  From now on we will ignore the bottom totally -- it never changes.  The top has two PRODUCTS, so we use PRODUCT-TO-SUM for each term.  You should check my algebra in all spots.

   k+1        k
sin(---x) sin(---x) + sin((k+1)x) sin(x/2) =
    2         2           
1  [      1          2k+1         2k+1         2k+3   ]
--- [ cos(---x) - cos(----x) + cos(----x) - cos(----x) ] =
2  [      2            2            2            2    ]


1  [      1                                    2k+3   ]
--- [ cos(---x)                           - cos(----x) ] =
2  [      2                                      2    ]

Now we have a difference of two cosines and we use the second formula:
1  [         2k+4       -2k-2   ]
--- [ - 2 sin(----x) sin(-----x) ]  =
2  [           4          4     ]

       k+2       -k-1
-  sin(---x) sin(----x)  
        2          2

We're close.  Just remember that sin(-A) = - sin A and we have:

     k+2       k+1
+ sin(---x) sin(---x)  
      2         2

AND THAT'S IT!  THAT'S THE NUMERATOR OF THE RIGHT HAND SIDE OF THE STATEMENT TO BE PROVED.  (Hooray!)