I have a question, here it is -
Given $Y = \sin(\theta)$, find percentage error in $y$ if percentage error in $\theta$ is $2\%$ at $\theta = \dfrac{\pi}{6} \ \text{rad}$.
Our teacher did this solution -
$$y = \sin \theta.$$
Differentiate $y$ wrt $\theta$
$$\frac{\mathrm{d}y}{\mathrm{d}\theta} = \cos \theta \iff \mathrm{d}y = \cos \theta \, \mathrm{d}\theta.$$
Divide both sides by $y$:
$$\frac{\mathrm{d}y}{y} = \frac{\cos \theta \, \mathrm{d}\theta}{\sin \theta}.$$
Multiply by $100$ on both sides
$$\left(\frac{\mathrm{d}y}{y}\right) \times 100 = \frac{\cos \theta \, \mathrm{d}\theta}{\sin \theta} \times 100.$$
So:
$$ \% \ \text{error in} \, y = \cot \theta \ \mathrm{d}\theta \times 100.$$
Multiply and divide by $\theta$ on RHS:
$$ \% \ \text{error in} \, y = \frac{\cot \theta \ \mathrm{d}\theta \cdot \theta}{\theta} \times 100.$$
Putting the values of $$ \frac{\mathrm{d}\theta}{\theta} \times 100 = 2 \% \quad \text{and} \quad \theta = \frac{\pi}{6} \ \text{rad},$$
we have
$$\begin{align}
\% \ \text{error in} \, y &= \cot \frac{\pi}{6} \times 2 \times \frac{\pi}{6} \\
&= \sqrt 3 \times \frac{\pi}{3} \\
&= \frac{\pi}{\sqrt 3} \ \% \\
&\approx 1.81 \%.
\end{align}$$
But I did this:
$$y = \sin \theta$$
$$y_t = \sin \frac{\pi}{6} = \frac{1}{2} = 0.5,$$
where $y_t$ means true value of $y$.
Assuming the case of positive error:
$$\begin{align} y_m &= \sin \frac{\pi}{6} + \frac{2}{100} \times \frac{\pi}{6} \\ &= \sin \frac{\pi}{6} \cos \frac{\pi}{300} + \cos \frac{\pi}{6} \sin \frac{\pi}{300} \\ &\approx 1. \end{align}\\ $$ \begin{align} \therefore \, \% \ \text{error in} \, y &= \frac{|y_t - y_m|}{y_t} \times 100 \\ &= \frac{|0.5 - 1|}{0.5} \times 100 \\ &= 100\%, \end{align}
where $y_m$ means measured value of $y$.
I want to know what mistake I made or why my approach is wrong. Thank you.
