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I want to prove the orthogonality of the functions: $\sin\left(\dfrac{2\pi x}{b-a}\right)$ and $\cos\left(\dfrac{2\pi x}{b-a}\right)$, where $b=\pi$ and $a = e$

My work:

$$\begin{align} \int^{\pi}_{e} \frac{1}{2} \sin\left(\frac{4\pi x}{\pi - e}\right)dx &= -\frac{\pi - e}{8 \pi} \left[\cos\left(\frac{4\pi x}{\pi - e} \right)\right]^{\pi}_e \tag{1}\\[6pt] &= \frac{e-\pi}{8\pi}\left[\cos\left(\frac{4\pi^2}{\pi - e} \right) - \cos\left(\frac{4\pi e}{\pi - e}\right) \right] \tag{2}\\[6pt] &= \frac{\pi - e}{4\pi} \left[\sin\left(\frac{2\pi (\pi - e)}{\pi - e} \right)\sin\left(\frac{2\pi (\pi + e)}{\pi - e} \right) \right] \tag{3}\\[6pt] = 0 \end{align}$$

Haven't I made any mistake?

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    $\begingroup$ Step $(2)$ to $(3)$ is incorrect. You cannot factor the cosine arguments like that. (That is, $\cos(pq)-\cos(pr) \neq \cos(p)(q-r)$.) What you can do is apply the sum-to-product identity $$\cos a - \cos b = -2\sin\left(\frac{a+b}{2}\right)\sin\left(\frac{a-b}{2}\right)$$ Then you can see about simplifying the individual arguments within the sines. $\endgroup$ – Blue May 24 '20 at 22:32
  • $\begingroup$ @Blue Ok, now I corrected the third equation and I get that the functions are orthogonal. $\endgroup$ – user May 24 '20 at 22:43
  • $\begingroup$ That's it. Congratulations! :) $\endgroup$ – Blue May 24 '20 at 22:50
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No mistake.

In general, the two functions $f$ & $g$ are said to be orthogonal if the integral of their product (treating as dot or inner product of two vectors) over some arbitrary interval is zero
$$\langle f,g\rangle =\int f(x)g(x)dx=0$$

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