the definite integrals of various combinations of sine and cousins on the interv
ID: 2855464 • Letter: T
Question
the definite integrals of various combinations of sine and cousins on the interval [0,2pi] The definite integrals of various combinations of sine and cosine on the interval (0, 2m] exhibit a number of intereating patterns. For now these patterns are simply curiositios and a source of additional problems for practice, but the patterns are very important as the foundation for an applied topic, Fourier Series, that you may encounter in more advanced courses. These problems ask you to show that the definite integral on (0, 2n] of sin(mx) multiplied by almost any other combination of sin(nx) or cos(nx) is 0. The only nonzero value comes when sin/mx) is multiplied by itself. -2 a) Show that if m and n are integers with m a n, then sin mz sin nr dz0 Hint: Use the sum and difference formulas from trig for cos(mx+nx) and cos(mx-nx) and subtract them to get an equation for sin(mx) sin(nx). b) Show that if m and n are integers, then sin mx cos nz dr = 0, (Consider m = n and m z n.) Hint: Use the sum and difference formulas from trig for sin(mx+nix) and sin(mx-nx) and add them together to get an equation for sin(mx) cos(nx) for m * n. For m n you can substitute so that both trig functions are in terms of mx and you should know how to integrate that c) Show that if m. O is an integer, then sin mir sin rnrdz .Explanation / Answer
a)[0 to 2] sinmx sinnx dx
sinasinb=(1/2)[cos(a-b)-cos(a+b)]
[0 to 2] (1/2)[cos(mx-nx)-cos(mx+nx)] dx
(1/2)[0 to 2] [cos(x(m-n))-cos(x(m+n))] dx
=(1/2)[0 to 2] [(1/(m-n))sin(x(m-n)) - (1/(m+n))sin(x(m+n)) +c]
=(1/2)[(1/(m-n))sin(2(m-n)) - (1/(m+n))sin(2(m+n)) +c] -(1/2) [(1/(m-n))sin(0) - (1/(m+n))sin(0) +c]
=(1/2)[(1/(m-n))sin(2(m-n)) - (1/(m+n))sin(2(m+n))]
m,n are integers so m-n,m+n are integers and term inside sine function is multiple of 2, so sine functions=0
=(1/2)[(1/(m-n))0 - (1/(m+n))0]
=0
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b)
m not equal to n
[0 to 2] sinmx cosnx dx
sinacosb=(1/2)[sin(a+b)+sin(a-b)]
[0 to 2] (1/2)[sin(mx+nx)+sin(mx-nx)] dx
=[0 to 2] (1/2)[-(1/(m+n))cos(x(m+n)) -(1/(m-n))cos(x(m-n))]
=(1/2)[-(1/(m+n))cos(2(m+n)) -(1/(m-n))cos(2(m-n))] -(1/2)[-(1/(m+n))cos(0(m+n)) -(1/(m-n))cos(0(m-n))]
=(1/2)[-(1/(m+n))) -(1/(m-n))] -(1/2)[-(1/(m+n)) -(1/(m-n))]
=0
m equal to n
[0 to 2] sinmx cosnx dx
[0 to 2] sinmx cosmx dx
[0 to 2] (1/2)2sinmx cosmx dx
[0 to 2] (1/2)sin2mx dx
=[0 to 2] (-1/2)(1/(2m))(cos2mx)
=(-1/(4m))[(cos2m(2)) -(cos0)]
=(-1/(4m))[1-1]
=0
===================================
c)
[0 to 2] sinmx sinmx dx
sinasinb=(1/2)[cos(a-b)-cos(a+b)]
[0 to 2] (1/2)[cos(mx-mx)-cos(mx+mx)] dx
[0 to 2] (1/2)[cos(0)-cos(2mx)] dx
[0 to 2] (1/2)[1 -cos(2mx)] dx
=[0 to 2] (1/2)[x - (1/(2m))sin(2mx)]
=(1/2)[2 - (1/(2m))sin(2m*2)] -(1/2)[0 - (1/(2m))sin(0)]
=(1/2)[2 - 0] -0
=
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