Respuesta :
Answer:
Step-by-step explanation:
We are going to consider the inner product
[tex] (f,g) = \int_{0}^1 f(x) g(x) dx[/tex]
Consider the following trigonometric identity:
[tex]\cos(\alpha-\beta) - \cos(\alpha+\beta) = 2 \sin(\alpha)\sin(\beta)[/tex]
Suppose that [tex]j\neq k[/tex] consider the inner product and the identity:
[tex](\phi_k,\phi_j) = \int_{0}^{1}\sin(k\pi x)\sin(j\pi x)dx=\frac{1}{2}\int_{0}^{1}\cos((k-j)\pi x)-\cos((k+j)\pi x)dx = \frac{1}{2}\left.(\frac{\sin((k-j)\pi x)}{(k-j)\pi}- \frac{\sin((k+j)\pi x)}{(k+j)\pi})\right|_{0}^{1}= \frac{1}{2}(\frac{\sin((k-j)\pi)}{(k-j)\pi}-\frac{\sin((k+j)\pi)}{(k+j)\pi}[/tex]
Note that since k,j are integers then k-j and k+j are integers. Then, by properties of the sine function we have that [tex]sin((k-j)\pi) = \sin((k+j)\pi) = 0 = (\phi_k,\phi_j)[/tex]
Consider also the following identity:
[tex]\sin^2(x) = \frac{1-\cos(2x)}{2}[/tex]
Suppose that j=k. Then
[tex](\phi_k, \phi_k) = \int_{0}^{1}\sin^2(k\pi x) dx = \int_{0}^{1}\frac{1-\cos(2k\pi x)}{2}dx = \left.\frac{x}{2}-\frac{\sin(2k\pi x)}{4k\pi }\right|_{0}^{1} = \frac{1}{2}-\frac{\sin(2k\pi)}{4k\pi }[/tex]
Since k is an integer, then [tex]\sin(2k\pi)=0[/tex]. So [tex](\phi_k,\phi_k) =\frac{1}{2}[/tex]
This proves what's been asked in the question.
