Applications of Series — Question 2

Question 2 – Solution

Step 1: Check the linear approximation.

For $x=0.2$, the sine series is alternating with decreasing term magnitudes. Using only $x$ gives $$|R|\le\frac{0.2^3}{3!}=0.001333\ldots>5\times10^{-5},$$ so the degree-$1$ polynomial does not provide the requested guarantee.

Step 2: Use the cubic polynomial.

$$P_3(x)=x-\frac{x^3}{3!},\qquad P_3(0.2)=0.2-\frac{0.008}{6}=\boxed{0.1986666667}.$$

Step 3: Bound the first omitted term.

$$|R|\le\frac{0.2^5}{5!}=\frac{0.00032}{120}\approx2.67\times10^{-6}<5\times10^{-5}.$$ The alternating sign gives $0.1986666666<\sin(0.2)<0.1986693334$; both endpoints round to $0.1987$. Therefore the smallest qualifying odd polynomial is cubic, and $\boxed{\sin(0.2)\approx0.1987}$ to four decimal places.