Show that the functions sin x, sin 2x, sin 3x, are orthogonal

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SUMMARY

The functions sin x, sin 2x, sin 3x, etc., are orthogonal on the interval (0, π) with respect to the weight function p(x) = 1. The orthogonality can be established using the integral 0π sin(nx) sin((n+1)x) dx = 0 for positive integers n. A recommended approach involves using the identity sin(s) sin(t) = (cos(s-t) - cos(s+t))/2 or expressing the sine functions as complex exponentials, sin(nx) = (einx - e-inx)/(2i), which simplifies the integration process.

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Show that the functions sin x, sin 2x, sin 3x, ... are orthogonal on the interval (0,pi) with respect to p(x) = 1 (where p is supposed to be rho)

i know i have to use this
\int_{0}^{\pi} \phi (x) \ psi (x) \rho (x) dx = 0 and i have no trouble doing it for n = 1, and n=2
but how wouldi go about proving it for hte general case that is

\int_{0}^{pi} \sin{nx} \sin{(n+1)x} dx for n in positive integers

would i use this identity :
\sin{s) \sin{t} = \frac{ \cos{s-t} - \cos{s+t} }{2}
i proved it using this identity... but isn't this identity a bit too obscure? Isnt there a less 'weird' way of doing this?
 
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One very easy way of proceeding is to write
<br /> \sin{(n x)} = \frac{e^{i n x} - e^{-inx} }{2i},<br />
expand both the sines under the integral as complex exponentials and the result falls out immediately.
 

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