Or search by topic
Aleksander from Gdynia Bilingual High School No 3, Poland used the properties of the sine function to find a polynomial approximation.
Andrei from Romania used the Taylor series and drew graphs to show the polynomial approximations .
To solve this problem, I use the Taylor series expansion around the origin. I see that the derivatives of $f(x) = \sin x $ are: $$\eqalign{ f^{4k}(x) &= \sin x \cr f^{4k+1}(x) &= \cos x \cr f^{4k+2}(x) &= -\sin x \cr f^{4k+3}(x) &= -\cos x }.$$ The formula for the Taylor expansion is: $$f(x) = f(x_0) + {(x - x_0)\over 1!}f'(x_0)+ {(x - x_0)^2\over 2!}f''(x_0)+... + {(x - x_0)^k\over k!}f^{(k)}(x_0)+ ...$$ Evidently, the odd-order derivatives are 0 for $x_0=0$. So, $\sin x$ could be written around the origin as: $$\sin x = x - {x^3\over 3!} + {x^5\over 5!} - {x^7\over 7!} + ...$$ The simplest method to test the accuracy of the series expansion is to represent on the same graph the function and its different order expansions.