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Summation Formulas for Trigonometric Functions


Date: 04/24/2001 at 11:21:02
From: Aaron Matthews
Subject: Summation formulas for trigonometric functions

How do you find the area under the curve of a trigonometric function 
using the definition of a limit and not an approximation? Are there 
summation formulas for trigonometric functions? If so, what are they 
and where did they come from?  

Thanks, Doctor Math.
Aaron


Date: 04/25/2001 at 13:55:55
From: Doctor Rob
Subject: Re: Summation formulas for trigonometric functions

Thanks for writing to Ask Dr. Math, Aaron.

There are such formulas, based on the following identities:

   sin(x) + sin(y) =  2*sin[(x+y)/2]*cos[(x-y)/2]
   sin(x) - sin(y) =  2*cos[(x+y)/2]*sin[(x-y)/2]
   cos(x) + cos(y) =  2*cos[(x+y)/2]*cos[(x-y)/2]
   cos(x) - cos(y) = -2*sin[(x+y)/2]*sin[(x-y)/2]

which are derived from the identities for sine and cosine of a sum
or difference of angles.

This will work best if you are evaluating the integral by taking a
partition of the interval into 2^n equal subintervals, evaluating
the function at the midpoint of each subinterval, and then taking
the limit as n increases without bound. For example, to integrate
sin(x) over the interval [a,b], each subinterval will have length
L(n) = (b-a)/2^n, and you get the Riemann sum

               2^n
   S(n) = L(n)*SUM sin[a+(k-1/2)*L(n)].
               k=1

By combining the terms for k = 2*r-1 and 2*r for each r with
1 <= r <= 2^(n-1),

   sin[a+(2*r-3/2)*L(n)] + sin[a+(2*r-1/2)*L(n)]
      = 2*sin[a+(2*r-1)*L(n)]*cos[L(n)/2]
      = 2*sin[a+(r-1/2)*L(n-1)]*cos[L(n+1)]

you get

             2^(n-1)
   S(n) = L(n)*SUM 2*sin[a+(r-1/2)*L(n-1)]*cos[L(n+1)]
               r=1

                           2^(n-1)
        = L(n)*2*cos[L(n+1)]*SUM sin[a+(r-1/2)*L(n-1)]
                             r=1

Repeating this n-1 times will reduce the sum to a single term:

   S(n) = L(n)*2^n*cos[L(n+1)]*...*cos[L(3)]*cos[L(2)]*sin[(a+b)/2].

Now you can show by induction that for any i >= 1 and any t,

   cos[2^(i-1)*t]*...*cos[2*t]*cos[t] = 2^(-i)*sin[2^i*t]/sin[t].

Substitution with t = L(n+1) and i = n then gives

   S(n) = L(n)*2^n*2^(-n)*sin[L(1)]/sin[L(n+1)]*sin[(a+b)/2]
        = 2*sin[(b-a)/2]*sin[(b+a)/2]*L(n+1)/sin[L(n+1)]

Now as you take the limit,

   S =  lim S(n)
       n->oo

     = 2*sin[(b-a)/2]*sin[(b+a)/2]*lim L(n+1)/sin[L(n+1)]
                                  n->oo

     = 2*sin[(b-a)/2]*sin[(b+a)/2]*lim x/sin(x)
                                   x->0

     = 2*sin[(b-a)/2]*sin[(b+a)/2]

     = cos(a) - cos(b)

again using one of the identities given at the beginning.

I leave it to you to repeat this process for the integral of cos(x).

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
College Trigonometry
High School Calculus
High School Trigonometry

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