My linear algebra text discusses the solution to minimization problems by use of the Gram-Schmidt procedure, which produces the an approximation such as $$\sin{x} = \frac{105(1485 - 153\pi^2 + \pi^4)}{8\pi^6}x - \frac{315(1155 -125\pi^2 + \pi^4 )}{4\pi^8}x^3 + \frac{693(945 - 105 \pi^2 + \pi^4)}{8\pi^{10}}x^5$$
by minimization of the inner product $$\int_{-\pi}^{\pi} (f(x) - \sin{x} )^2 dx$$
I later read this article, and the algorithms discussed can easily be applied to the sine function. How is this procedure different from the one above?
The objective function is not the same; both methods can be applied but the one you mention is really simpler.
Let me try to explain why using simple words.
Suppose that you want to fit $\sin(x)$ between $-\pi$ and $\pi$ using a pentic polynomial without intercept and with only odd powers. You could generate a table of $n$ points $\Big(x_i,\sin(x_i)\Big)$ and proceed to a multilinear square fit procedure. If you consider an infinite number of data points, minimizing the sum of squares of residuals is equivalent to minimizing $$F=\int_{-\pi}^{\pi} (f(x) - \sin{x} )^2 dx$$ So, consider that $f(x)=a x+bx^3+cx^5$. Replacing and integrating between bounds leads to $$F=\frac{2 \pi ^3 a^2}{3}+\frac{2}{7} \pi ^7 \left(2 a c+b^2\right)-4 \pi \left(a+\left(\pi ^2-6\right) b+\left(120-20 \pi ^2+\pi ^4\right) c\right)+\frac{4}{5} \pi ^5 a b+\frac{4}{9} \pi ^9 b c+\frac{2 \pi ^{11} c^2}{11}+\pi$$ and now you want to minimize $F$; so you take the partial derivatives with respect to $a,b,c$ and you set them equal to $0$. The derivatives are given by $$\frac{dF}{da}=\frac{4 \pi ^3 a}{3}+\frac{4 \pi ^5 b}{5}+\frac{4 \pi ^7 c}{7}-4 \pi=0$$ $$\frac{dF}{db}=\frac{4 \pi ^5 a}{5}+\frac{4 \pi ^7 b}{7}+\frac{4 \pi ^9 c}{9}-4 \pi \left(\pi ^2-6\right)=0$$ $$\frac{dF}{dc}=\frac{4 \pi ^7 a}{7}+\frac{4 \pi ^9 b}{9}+\frac{4 \pi ^{11} c}{11}-4 \pi \left(120-20 \pi ^2+\pi ^4\right)=0$$ This makes a system of three linear equations in $a,b,c$ for which the solution is $$a=\frac{105 \left(1485-153 \pi ^2+\pi ^4\right)}{8 \pi ^6}$$ $$b=-\frac{315 \left(1155-125 \pi ^2+\pi ^4\right)}{4 \pi ^8}$$ $$c=\frac{693 \left(945-105 \pi ^2+\pi ^4\right)}{8 \pi ^{10}}$$ For these values, we find $$F=\frac{-19646550+4365900 \pi ^2-287280 \pi ^4+5040 \pi ^6-42 \pi ^8+\pi ^{10}}{\pi ^9} \simeq 0.000116169$$ If Taylor expansion had been used, the result would heve been $$F=-11 \pi +2 \pi ^3-\frac{\pi ^5}{6}+\frac{4 \pi ^7}{315}-\frac{\pi ^9}{1620}+\frac{\pi ^{11}}{79200} \simeq 0.118702$$ which is almost $1000$ times larger.
By the way, there is a typo in your formula or in your book : $603$ should be $693$.
Added later to this answer
Just for personal fun, I continued the same exercise adding the term $x^7$; for this case $F=1.54037*10^{-7}$. Adding the term $x^9$, $F=8.40998*10^{-11}$. This is really quite impressive.