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anonymous
 4 years ago
Show that for all integers m and n, with m ≠ +/n, the integral from π to π of cos(mθ)cos(nθ) dθ = 0
anonymous
 4 years ago
Show that for all integers m and n, with m ≠ +/n, the integral from π to π of cos(mθ)cos(nθ) dθ = 0

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myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1okay and we also know that \[\cos(m \theta  n \theta)=\cos(m \theta) \cos(n \theta)+\sin(m \theta) \sin(n \theta)\] so this means we have \[\sin(m \theta) \sin(n \theta)=\cos(m \thetan \theta)\cos(m \theta)\cos(n \theta)\] and remember that \[\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\sin(m \theta)\sin(n \theta) \] \[\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\cos(m \thetan \theta)\cos(m \theta)\cos(n \theta)\] \[2 \cos(m \theta) \cos(n \theta)= \cos(m \theta+n \theta)+\cos(m \thetan \theta)\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1we are almost there now

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\cos(m \theta) \cos(n \theta)= \frac{1}{2} (\cos(m \theta+n \theta)+\cos(m \thetan \theta))\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so this means we have \[\frac{1}{2}\int\limits_{\pi}^{\pi}(\cos([m+n] \theta)+\cos([mn] \theta) d \theta\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[[\frac{1}{2}\frac{1}{m+n} \sin([m+n] \theta)+\frac{1}{2} \frac{1}{mn}\sin([mn] \theta)]_{\pi}^{\pi}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1i will assume m and n are integers

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1this means m+n is an integer and mn is an integer sin(integer * pi) is 0 sin(integer * (pi)) is 0 m+n can't be zero mn can't be zero so the answer is 0 as long as m does not equal n or m equals n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in first part didnt u forget one sin(mt)sin(nt) ?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1? go to the your post and read what i posted there

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1i put some more steps in the one i posted for you

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e72a5830b8b247045cb6cde
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