Compute approximately: 1° sin 46.3^circ; 2° cos
Compute approximately: 1° $\sin 46.3^{\circ}$; 2° $\cos 59.1^{\circ}$; 3° $\sin 1^{\circ}$.
Explanation of Method
Small steps from standard angles — with the degree-to-radian factor kept in view.
Step 1: The linearisation, with the conversion factor
Derivative formulas for trigonometric functions hold only in radian measure, so a step of $\Delta\varphi$ degrees must be converted before it multiplies the derivative: \[f(\varphi) \approx f(\varphi_{0})+f'(\varphi_{0})\cdot\Delta\varphi\cdot\frac{\pi}{180}. \tag{1}\]
Omitting the factor $\pi/180 \approx 0.017453$ is by far the most common error in problems of this type.
Step 2: Part 1°: $\sin 46.3^{\circ}$
Base point $45^{\circ}$, step $\Delta\varphi = 1.3^{\circ}$, and $\sin45^{\circ} = \cos45^{\circ} = \dfrac{\sqrt2}{2} \approx 0.70711$: $1.3\cdot\frac{\pi}{180} \approx 0.022689$.
\[ \sin46.3^{\circ} \approx 0.70711+0.70711\times0.022689 = 0.70711\bigl(1.022689\bigr) \approx 0.72315. \]
Step 3: Part 2°: $\cos 59.1^{\circ}$
Base point $60^{\circ}$, step $\Delta\varphi = -0.9^{\circ}$. Since $\bigl(\cos\varphi\bigr)' = -\sin\varphi$, the two minus signs cancel: \[\cos59.1^{\circ} \approx \cos60^{\circ}+\sin60^{\circ}\cdot0.9\cdot\frac{\pi}{180} = 0.5+0.86603\times0.015708,\]
\[ \cos59.1^{\circ} \approx 0.5+0.013603 = 0.51360. \]
Step 4: Part 3°: $\sin 1^{\circ}$
Base point $0^{\circ}$, where $\sin0 = 0$ and $\cos0 = 1$, so the linearisation is nothing but the conversion factor itself: $\sin1^{\circ} \approx 1\cdot1\cdot\frac{\pi}{180} = \frac{\pi}{180} \approx 0.0174533$.
Step 5: Verification
\[\begin{array}{c|c|c|c} & \text{estimate} & \text{exact} & \text{error} \\ \hline \sin46.3^{\circ} & 0.7231505 & 0.7229671 & +1.8\cdot10^{-4} \\ \cos59.1^{\circ} & 0.5136035 & 0.5135413 & +6.2\cdot10^{-5} \\ \sin1^{\circ} & 0.0174533 & 0.0174524 & +8.9\cdot10^{-7} \end{array}\]
\[\text{Second-order check, part 1}^{\circ}: \ \tfrac12\sin45^{\circ}\bigl(0.022689\bigr)^{2} = 1.82\cdot10^{-4} \;\;\checkmark\]
\[\text{Part 3}^{\circ}\text{ error } \approx \frac{1}{6}\left(\frac{\pi}{180}\right)^{3} = 8.9\cdot10^{-7} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\sin46.3^{\circ}\approx0.7232, \qquad \cos59.1^{\circ}\approx0.5136, \qquad \sin1^{\circ}\approx\frac{\pi}{180}\approx0.01745\,}\]
Remark
Remark. All three estimates come out high, and the middle column of the table explains the pattern of sizes: the error is $-\tfrac12 f''(\varphi_{0})h^{2}$ with $h$ the step in radians, so it scales as the square of the step. Part 3° is exceptional — there $f''(0) = -\sin0 = 0$, the quadratic term vanishes entirely, and the first correction is cubic, which is why $\sin1^{\circ} = \pi/180$ is accurate to six decimals.