## pricetag 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

1. myininaya

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 $\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\sin(m \theta)\sin(n \theta)$ $\cos(m \theta) \cos(n \theta)=\cos(m \theta+n \theta)+\cos(m \theta-n \theta)-\cos(m \theta)\cos(n \theta)$ $2 \cos(m \theta) \cos(n \theta)= \cos(m \theta+n \theta)+\cos(m \theta-n \theta)$

2. myininaya

we are almost there now

3. myininaya

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

4. myininaya

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

5. myininaya

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

6. myininaya

i will assume m and n are integers

7. myininaya

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

8. myininaya

:)

9. simagholami

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

10. myininaya

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

11. myininaya

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

12. myininaya