Question

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

Answer 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
\[\sin(m \theta) \sin(n \theta)=\cos(m \theta-n \theta)-\cos(m \theta)\cos(n \theta)\]

and remember that
\[\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\sin(m \theta)\sin(n \theta) \]

\[\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\cos(m \theta-n \theta)-\cos(m \theta)\cos(n \theta)\]

\[2 \cos(m \theta) \cos(n \theta)= \cos(m \theta+n \theta)+\cos(m \theta-n \theta)\]

Answer 2

we are almost there now

Answer 3

\[\cos(m \theta) \cos(n \theta)= \frac{1}{2} (\cos(m \theta+n \theta)+\cos(m \theta-n \theta))\]

Answer 4

so this means we have
\[\frac{1}{2}\int\limits_{-\pi}^{\pi}(\cos([m+n] \theta)+\cos([m-n] \theta) d \theta\]

Answer 5

\[[\frac{1}{2}\frac{1}{m+n} \sin([m+n] \theta)+\frac{1}{2} \frac{1}{m-n}\sin([m-n] \theta)]_{-\pi}^{\pi}\]

Answer 6

i will assume m and n are integers

Answer 7

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

Answer 8

:)

Answer 9

in first part didnt u forget one sin(mt)sin(nt) ?

Answer 10

?
go to the your post and read what i posted there

Answer 11

i put some more steps in the one i posted for you

Answer 12

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e72a5830b8b247045cb6cde