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Show that for all integers m and n, with m ≠ +/n, the integral from π to π of cos(mθ)cos(nθ) dθ = 0
 2 years ago
 2 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
 2 years ago
 2 years ago

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myininayaBest ResponseYou've already chosen the best response.1
okay 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 \[\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\sin(m \theta)\sin(n \theta) \] \[\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\cos(m \thetan \theta)\cos(m \theta)\cos(n \theta)\] \[2 \cos(m \theta) \cos(n \theta)= \cos(m \theta+n \theta)+\cos(m \thetan \theta)\]
 2 years ago

myininayaBest ResponseYou've already chosen the best response.1
we are almost there now
 2 years ago

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

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

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

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

myininayaBest ResponseYou've already chosen the best response.1
this 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
 2 years ago

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

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

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

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