Deduce orthogonality relations for sine and cosine w/ Euler's Formula

[zenterix]

Homework Statement

The following is problem 7 of chapter 9.10 in Apostol's *Calculus*, Volume I.

(a) If $m$ and $n$ are integers prove that

$$\int_0^{2\pi} e^{inx}e^{-imx}dx=\begin{cases}0\ \ \ \ \ \text{if}\
m\neq n\\ 2\pi\ \ \ \ \text{if}\ m=n\end{cases}$$

(b) Use part (a) to deduce the orthogonality relations for the sine and
cosine ($m$ and $n$ are integers, $m^2\neq n^2$):

$$\int_0^{2\pi} \sin{nx}\cos{mx}dx=\int_0^{2\pi}
\sin{nx}\sin{mx}dx=\int_0^{2\pi} \cos{nx}\cos{mx}dx$$

Relevant Equations

$$\int_0^{2\pi} \sin^2{nx}dx=\int_0^{2\pi} \cos^2{nx}dx=\pi$$

To solve part (a), we write $e^{inx}e^{-imx}=e^{ix(n-m)}$.

If $m=n$ then this expression is 1, and so the integral of 1 from 0 to $2\pi$ is $2\pi$.

If $m\neq n$ then we use Euler's formula and integrate. The result is zero.

My question is how do we solve part (b) using part (a)?

I can solve part (b) by using the trigonometric identities for $\sin{(a\pm b)}$ and $\cos{(a\pm b)}$ with $a=mx$ and $b=nx$.

But how do we solve (b) using part (a)?

What I tried to do was

$$e^{nxi}e^{-mxi}=(\cos{nx}+i\sin{mx})(\cos{mx}-i\sin{nx})\tag{1}$$

$$e^{nxi}e^{-mxi}=\cos{nx}\cos{mx}+\sin{nx}\sin{mx}+i(-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})\tag{2}$$

If $m\neq n$ then when we integrate this expression the left-hand side is zero (by part (a)).

However, we are left with

$$0=\int_0^{2\pi} (\cos{nx}\cos{mx}+\sin{nx}\sin{mx}) dx + i\int_0^{2\pi} (-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})dx$$

Thus

$$\int_0^{2\pi} (\cos{nx}\cos{mx}+\sin{nx}\sin{mx}) dx=0$$

$$\int_0^{2\pi} (-\cos{nx}\sin{mx}+\sin{nx}\cos{mx})dx=0$$

 

[FactChecker]

Converting the exponentials to sin and cos is the wrong way to go. Instead, convert the sin and cos to exponentials.