Summation Formulas for Trigonometric FunctionsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/