Question

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

Answer 1

\[\cos p \sin q=\frac{1}{2}(\sin(p+q)-\sin(p-q))\]

Answer 2

use that and then integrate

Answer 3

so 1/2(sin(mθ+nθ)-sin(mθ+nθ))
derived would be
(1/2)(cos(mθ+nθ)(mθdθ)(nθdθ)-cos(mθ+nθ)(mθdθ)(nθdθ)) is that right?

Answer 4

and then plug in π for θ minus the quantity of -π for θ

Answer 5

emm....u must integrate :/

Answer 6

hah oops

Answer 7

=\[\int\limits_{-π}^{π}sin(mθ+nθ)]-\int\limits_{-π}^{π}sin(mθ-nθ)\]

Answer 8

yes integrate and apply bounds

Answer 9

one another point...integrand is an odd function and ur interval is symmetric with respect to y-axis

Answer 10

i honestly don't even know where to begin integrating those. i suck at trig integrals

Answer 11

nevermind, i figured it out, thanks!