## Problem Statement

Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.

\[
\left\{
\begin{array}{l}
x=\sin t \\
y=\cos 2 t
\end{array}
\right.
\]

$t_{0}=\frac{\pi}{6}$
## Solution

Since $t_{0}=\frac{\pi}{6}$, then

$x_{0}=\sin \frac{\pi}{6}=0.5$

$y_{0}=\cos \left(2 \cdot \frac{\pi}{6}\right)=\cos \frac{\pi}{3}=0.5$

Let's find the derivatives:

$x_{t}^{\prime}=(\sin t)^{\prime}=\cos t$

$y_{t}^{\prime}=(\cos 2 t)^{\prime}=-2 \sin 2 t$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{-2 \sin 2 t}{\cos t}=-\frac{4 \sin t \cdot \cos t}{\cos t}=-4 \sin t$

Then:

$y_{0}^{\prime}=-4 \sin \frac{\pi}{6}=-4 \cdot 0.5=-2$

Equation of the tangent line:
$y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$

$y-0.5=-2 \cdot(x-0.5)$

$y=-2 x+1+0.5$

$y=-2 x+1.5$

Equation of the normal line:

$$
\begin{aligned}
& y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right) \\
& y-0.5=-\frac{1}{-2} \cdot(x-0.5) \\
& y=\frac{x}{2}-0.25+0.5 \\
& y=\frac{x}{2}+0.25
\end{aligned}
$$

## Problem Kuznetsov Differentiation 17-30