Prove the following approximate formulas for sufficiently
Prove the following approximate formulas for sufficiently small $\alpha$: 1° $\cos\alpha\approx1$; 2° $\sin\alpha\approx\alpha$; 3° $\tan\alpha\approx\alpha$.
Explanation of Method
The small-angle formulas, with the orders of their errors.
Step 1: Part 1°: $\cos\alpha\approx1$
\[ f(\alpha) = \cos\alpha, \qquad f(0) = 1, \qquad f'(\alpha) = -\sin\alpha, \qquad f'(0) = 0. \]
The linear term vanishes, so the linearisation is simply the constant $1$: $\cos\alpha \approx 1+0\cdot\alpha = 1$.
Since $f''(0) = -1$, the expansion is $\cos\alpha = 1-\dfrac{\alpha^{2}}{2}+O(\alpha^{4})$, and the error is $O(\alpha^{2})$.
Step 2: Part 2°: $\sin\alpha\approx\alpha$
\[ f(\alpha) = \sin\alpha, \qquad f(0) = 0, \qquad f'(\alpha) = \cos\alpha, \qquad f'(0) = 1, \]
\[ \sin\alpha \approx 0+1\cdot\alpha = \alpha. \]
Here $f''(0) = 0$ as well, so the quadratic term is absent and
\[ \sin\alpha = \alpha-\frac{\alpha^{3}}{6}+O\bigl(\alpha^{5}\bigr), \]
giving an error of only $O(\alpha^{3})$ — one order better than a generic linearisation.
Step 3: Part 3°: $\tan\alpha\approx\alpha$
\[ f(\alpha) = \tan\alpha, \qquad f(0) = 0, \qquad f'(\alpha) = \frac{1}{\cos^{2}\alpha}, \qquad f'(0) = 1, \]
\[ \tan\alpha \approx \alpha. \]
Step 4: The error in Part 3°
Dividing the two expansions: \[\tan\alpha = \frac{\alpha-\dfrac{\alpha^{3}}{6}+O(\alpha^{5})}{1-\dfrac{\alpha^{2}}{2}+O(\alpha^{4})} = \left(\alpha-\frac{\alpha^{3}}{6}\right)\left(1+\frac{\alpha^{2}}{2}+O(\alpha^{4})\right)+\cdots\]
\[ = \alpha+\frac{\alpha^{3}}{2}-\frac{\alpha^{3}}{6}+O\bigl(\alpha^{5}\bigr) = \alpha+\frac{\alpha^{3}}{3}+O\bigl(\alpha^{5}\bigr). \]
So the error is $+\dfrac{\alpha^{3}}{3}$, again $O(\alpha^{3})$ but twice the size of the sine error and of the opposite sign.
Step 5: Verification at $\alpha = 0.1\ \text{rad}$
\[\begin{array}{c|c|c|c|c} \text{formula} & f(0) & f'(0) & \text{error order} & \text{value at } 0.1 \\ \hline \cos\alpha\approx1 & 1 & 0 & O(\alpha^{2}) & 0.9950042 \text{ vs } 1 \\ \sin\alpha\approx\alpha & 0 & 1 & O(\alpha^{3}) & 0.0998334 \text{ vs } 0.1 \\ \tan\alpha\approx\alpha & 0 & 1 & O(\alpha^{3}) & 0.1003347 \text{ vs } 0.1 \end{array}\]
\[\text{Errors: } -0.0049958 \approx -\frac{\alpha^{2}}{2};\quad -0.0001666 \approx -\frac{\alpha^{3}}{6};\quad +0.0003347 \approx +\frac{\alpha^{3}}{3} \;\;\checkmark\]
\[\text{At } \alpha = 1^{\circ} = 0.0174533:\ \sin\alpha = 0.0174524,\ \tan\alpha = 0.0174551 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\cos\alpha\approx1, \qquad \sin\alpha\approx\alpha, \qquad \tan\alpha\approx\alpha\,}\]
Remark
Remark. These formulas require $\alpha$ in radians; in degrees $\sin\alpha\approx\alpha$ is wildly false. The sine and tangent share the same linearisation because both have $f(0)=0$ and $f'(0)=1$; they first part company at order $\alpha^{3}$, where the sine falls below the line and the tangent rises above it. This is the analytic content of the classical inequality $\sin\alpha\lt \alpha\lt \tan\alpha$ for $0\lt \alpha\lt \tfrac{\pi}{2}$.