Write down an expression (involving inner products) for the values of the constants in the general solution. In the process of finding the general solution to a partial differential equation using separation of variables, I derived the Fourier sine series $\sum_{n = 1}^\infty b_n \sin(n\pi x) = e^{2x}$.
The problem is as follows:

Write down an expression (involving inner products) for the values of the constants in the general solution.

The solution provided is as follows:

The initial condition is $u(x, 0) = 1$ for $0 < x < 1$.
$\therefore \sum_{n = 1}^\infty b_n e^{-2x} \sin(n\pi x) = 1$
$\implies \sum_{n = 1}^\infty b_n \sin(n\pi x) = e^{2x}$
The functions $\sin(n\pi x)$, $n = 1, 2, 3, ...$ are orthogonal on $[0, 1]$ with inner product $\langle\ f(x), g(x)\rangle = \int_0^1 f(x)g(x) \ dx$, and hence
$b_n \dfrac{\langle e^{2x}, \sin(n\pi x) \rangle}{\langle \sin(n\pi x), \sin(n\pi x) \rangle}$.

I don't understand how it was found that the functions $\sin(n\pi x)$, $n = 1, 2, 3, ...$ are orthogonal on $[0, 1]$?
I also don't understand how we get $b_n \dfrac{\langle e^{2x}, \sin(n\pi x) \rangle}{\langle \sin(n\pi x), \sin(n\pi x) \rangle}$? It seems almost like an orthonormalisation?
I would greatly appreciate it if people could please take the time to clarify this.
 A: You can convince yourself that they are indeed orthogonal by checking the integral :
$$\int_0^1 \sin(n \pi x) \sin(m \pi x) = \frac{1}{2}\delta_{m-n}  $$
For your second question, it is exactly an orthonormalization. The inner product is equivalent to a dot product of ordinary vectors. The vector basis are given by $e_n = \sin(n\pi x)$, which is simply an infinite basis. You are simply doing projections into this vector basis. Consider the following : $$ \exp(2x) = \sum_n b_n \sin(n\pi x)$$. Now integrate both sides with another sine function and permute sum/integral, which is how you usually derive $b_n$: $$ \int_0^1 \exp(2x) \sin(m \pi x) = \sum_n b_n \int_0^1  \sin(m\pi x) \sin(n\pi x)$$. Using the fact that the right integral is zero if $n\neq m$, you get: $$ \int_0^1 \sin(m \pi x )\exp(2x) = b_m \int_0^1 \sin(m\pi x)^2$$. This is exactly a vector base change. The sines function form a vector space, with the inner product defined by an integral. The vector space is also complete in the sense that any function in [0,1] can be exactly represneted by a sum of sines. 
