Proving the orthogonality of $\sin\frac{2\pi x}{\pi-e}$ and $\cos\frac{2\pi x}{\pi-e}$ 
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?

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
$$\langle f,g\rangle =\int f(x)g(x)dx=0$$
