cubic approximation with four points(approximating sine function with polynomials)

Question

I was reading the following article https://mixedmath.wordpress.com/2013/11/17/an-intuitive-overview-of-taylor-series/ regarding Taylor Series.When I got to the part 1.3. Cubic approximation I got lost. I don't get how p3 is equal to all that.

Answer 1

You want the second paragraph in that section, he explains it there. He is trying to use the points $-\pi/3, -\pi/6, \pi/6,$ and $\pi/3$ to generate $p_3(x)$, a cubic approximating the sine.

Find a polynomial-part that takes the right value at $-\pi/3$ and is $0$ at $-\pi/6, \pi/6$ and $\pi/3$. Do the same for the other three points. Add these together. This is reasonable since it’s easy to find a cubic that’s $0$ at the points $-\pi/6, \pi/6$ and $\pi/3$: it’s $$ a(x+\pi/6)(x - \pi/6)(x-\pi/3). $$

We want to choose the value of $a$ so that this piece is $\sin(-\pi/3) = -\sqrt{3}/2$ when $x=-\pi/3$. This leads us to choosing $$ a = \frac{-\sqrt{3}/2}{(-\pi/3 + \pi/6)(-\pi/3 - \pi/6)(-\pi/3 - \pi/3)}. $$