Derive the orthogonality relations on $[-p,p]$ for the functions $\cos\left(\frac{\pi mx}{p}\right),\sin\left(\frac{\pi nx}{p}\right)$using the trig identities

$\displaystyle \sin(u)\sin(v) = \frac{1}{2}\left[\cos(u-v)-\cos(u+v)\right]$

$\displaystyle \cos(u)\cos(v) = \frac{1}{2}\left[\cos(u-v)+\cos(u+v)\right]$

$\displaystyle \sin(u)\cos(v) = \frac{1}{2}\left[\sin(u+v)+\sin(u-v)\right]$

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$\displaystyle \int_{-p}^p \sin\!\left(\frac{\pi nx}{p}\right) \cos\!\left(\frac{\pi mx}{p}\right) dx$

$\displaystyle =\frac{1}{2}\int_{-p}^{p}\sin\Big($ $\Big)+\sin\Big($ $\Big)\hspace{5pt}dx$

The first function in the integrand has period while the second has period . The length of the interval $\lbrack -p ,p\rbrack$ is 2p, therefore the integral is



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$\displaystyle \int_{-p}^p \sin\!\left(\frac{\pi mx}{p}\right) \sin\!\left(\frac{\pi nx}{p}\right) dx$

$\displaystyle =\frac{1}{2}\int_{-p}^{p}\cos\Big($ $\Big)-\cos\Big($$\Big)\hspace{5pt}dx$

The first function in the integrand has period while the second has period . The length of the interval $\lbrack -p ,p\rbrack$ is 2p, therefore the integral is



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$\displaystyle \int_{-p}^p \cos\!\left(\frac{\pi mx}{p}\right) \cos\!\left(\frac{\pi nx}{p}\right) dx$

$\displaystyle =\frac{1}{2}\int_{-p}^{p}\cos\Big($ $\Big)+\cos\Big($ $\Big)\hspace{5pt}dx$

The first function in the integrand has period while the second has period . The length of the interval $\lbrack -p ,p\rbrack$ is 2p, therefore the integral is