Determine the interval for x such that the approximation

Determine the interval for $x$ such that the approximation $\sin x\approx x-\dfrac{x^3}{3!}=x-\dfrac{x^3}6$ has a relative error less than $1\%$.

Step 1: Identify the remainder

From $\sin x=x-\tfrac{x^3}6+\tfrac{x^5}{120}-\cdots$, the remainder after the cubic term is $R_3(x)=\sin x-(x-\tfrac{x^3}6)$; by the Lagrange form (the $x^4$ coefficient of $\sin x$ is zero, so the effective remainder sits at order 5):

\[ R_3(x) = \frac{\cos\xi}{120}x^5, \qquad |R_3(x)| \le \frac{x^5}{120}. \tag{1} \]

Step 2: Relative-error condition

Bounding the denominator conservatively by $\sin x\approx x$: \[\frac{x^5/120}{\sin x} \lt 0.01 \;\Longrightarrow\; \frac{x^4}{120} \lt 0.01 \;\Longrightarrow\; x \lt (1.2)^{1/4} \approx 1.0466. \tag{2}\]

Step 3: Sharper bound using $\sin x>x-x^3/6$

Replacing the denominator by the tighter lower bound $x-x^3/6$ and setting $u=x^2$: \[\frac{x^4/120}{1-x^2/6} \lt 0.01 \;\Longrightarrow\; u^2+0.2u-1.2\lt 0 \;\Longrightarrow\; u \lt \frac{-0.2+\sqrt{4.84}}2 = 1.0. \tag{3}\]

Hence $x^2\lt 1$, i.e.\ $|x|\lt 1$.

Step 4: Verification

\[x=1\text{: }\sin1\approx0.84147,\ 1-\tfrac16\approx0.83333,\ \text{relative error}\approx0.00967\lt 0.01\ \checkmark\]