Applications of Series — Question 8

Question 8 – Solution

Step 1: Use the local series.

$$\sin\theta=\theta-\frac{\theta^3}{6}+\frac{\theta^5}{120}-\cdots.$$ Thus the leading absolute error in replacing $\sin\theta$ by $\theta$ is approximately $\theta^3/6$, and the leading relative error is approximately $\theta^2/6$.

Step 2: Evaluate the exact relative error.

At $\theta=0.2$ radians, $$\sin(0.2)\approx0.1986693308,\qquad 0.2-\sin(0.2)\approx0.0013306692.$$ Therefore $$\frac{0.2-\sin(0.2)}{\sin(0.2)}\approx0.0066979=\boxed{0.6698\%}.$$

Step 3: Compare with the leading prediction.

$$\frac{\theta^2}{6}=\frac{0.04}{6}\approx0.0066667=0.6667\%,$$ which is close to the exact relative error. This quantifies why the approximation is useful but not exact.