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?

  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.