Compute approximately: 1° arcsin 0.4983; 2° arctan 1.05

Compute approximately: 1° $\arcsin 0.4983$; 2° $\arctan 1.05$.

Step 1: The linearisation, with the conversion factor

The derivative of an inverse trigonometric function delivers radians, so expressing the answer in degrees requires the reciprocal factor of the previous problem: \[f(x) \approx f(x_{0})+f'(x_{0})\bigl(x-x_{0}\bigr)\cdot\frac{180}{\pi} \quad [\text{degrees}], \qquad \frac{180}{\pi} \approx 57.29578. \tag{1}\]

Step 2: Part 1°: base point and step

Since $\sin30^{\circ} = 0.5$, take $x_{0} = 0.5$ with $\arcsin x_{0} = 30^{\circ}$ and

\[ \Delta x = 0.4983-0.5 = -0.0017. \]

\[ \bigl(\arcsin x\bigr)' = \frac{1}{\sqrt{1-x^{2}}}, \qquad f'(0.5) = \frac{1}{\sqrt{0.75}} = \frac{2}{\sqrt3}. \]

Step 3: Part 1°: the correction

The derivative and the conversion factor combine into a single constant: \[\frac{2}{\sqrt3}\cdot\frac{180}{\pi} = \frac{360}{\pi\sqrt3} = \frac{360}{5.441398} \approx 66.1595 \ \text{degrees per unit of } x.\]

\[ 66.1595\times\bigl(-0.0017\bigr) \approx -0.11247^{\circ}, \]

\[ \arcsin0.4983 \approx 30^{\circ}-0.11247^{\circ} = 29.8875^{\circ} = 29^{\circ}53.2'. \]

Step 4: Part 2°: base point and step

Since $\tan45^{\circ} = 1$, take $x_{0} = 1$ with $\arctan x_{0} = 45^{\circ}$ and $\Delta x = 0.05$: $\bigl(\arctan x\bigr)' = \frac{1}{1+x^{2}}, \qquad f'(1) = \frac12$.

Step 5: Part 2°: the correction

\[ \frac12\cdot0.05\cdot\frac{180}{\pi} = \frac{9}{2\pi} = \frac{9}{6.283185} \approx 1.4324^{\circ}, \]

\[ \arctan1.05 \approx 45^{\circ}+1.4324^{\circ} = 46.4324^{\circ} = 46^{\circ}25.9'. \]

Step 6: Verification

\[\arcsin0.4983 = 29.887592^{\circ} \text{ exactly; estimate } 29.887529^{\circ}, \text{ error } 6\cdot10^{-5}\,^{\circ} \;\;\checkmark\]

\[\arctan1.05 = 46.397181^{\circ} = 46^{\circ}23.8' \text{ exactly; estimate } 46.432394^{\circ}\]

\[\text{Error } 0.0352^{\circ} \approx 2.1' \text{ — the estimate overshoots by about two minutes of arc} \;\;\checkmark\]