Here's the question you clicked on:
pricetag
Show that for all integers m and n, with m ≠ +/-n, the integral from -π to π of cos(mθ)cos(nθ) dθ = 0
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 \[\sin(m \theta) \sin(n \theta)=\cos(m \theta-n \theta)-\cos(m \theta)\cos(n \theta)\] and remember that \[\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\sin(m \theta)\sin(n \theta) \] \[\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\cos(m \theta-n \theta)-\cos(m \theta)\cos(n \theta)\] \[2 \cos(m \theta) \cos(n \theta)= \cos(m \theta+n \theta)+\cos(m \theta-n \theta)\]
we are almost there now
\[\cos(m \theta) \cos(n \theta)= \frac{1}{2} (\cos(m \theta+n \theta)+\cos(m \theta-n \theta))\]
so this means we have \[\frac{1}{2}\int\limits_{-\pi}^{\pi}(\cos([m+n] \theta)+\cos([m-n] \theta) d \theta\]
\[[\frac{1}{2}\frac{1}{m+n} \sin([m+n] \theta)+\frac{1}{2} \frac{1}{m-n}\sin([m-n] \theta)]_{-\pi}^{\pi}\]
i will assume m and n are integers
this means m+n is an integer and m-n is an integer sin(integer * pi) is 0 sin(integer * (-pi)) is 0 m+n can't be zero m-n can't be zero so the answer is 0 as long as m does not equal -n or m equals n
in first part didnt u forget one sin(mt)sin(nt) ?
? go to the your post and read what i posted there
i put some more steps in the one i posted for you
http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e72a5830b8b247045cb6cde