## Problem Statement

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

\[
\left\{\begin{array}{l}
x=\sin t \\
y=\cos t
\end{array}\right.
\]
$t_{0}=\frac{\pi}{6}$
## Solution

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

$x_{0}=\sin \frac{\pi}{6}=\frac{1}{2}$

$y_{0}=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$

Let's find the derivatives:

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

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

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{-\sin t}{\cos t}=-\tan t$

Then:

$y_{0}^{\prime}=-\tan \frac{\pi}{6}=-\frac{1}{\sqrt{3}}$

Equation of the tangent line:

$y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$

$y-\frac{\sqrt{3}}{2}=-\frac{1}{\sqrt{3}} \cdot\left(x-\frac{1}{2}\right)$

$y=-\frac{1}{\sqrt{3}} \cdot x+\frac{1}{2 \sqrt{3}}+\frac{\sqrt{3}}{2}$

$y=-\frac{1}{\sqrt{3}} \cdot x+\frac{2}{\sqrt{3}}$

Equation of the normal line:

$y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right)$

$y-\frac{\sqrt{3}}{2}=-\frac{1}{\left(-\frac{1}{\sqrt{3}}\right)}\left(x-\frac{1}{2}\right)$

$y-\frac{\sqrt{3}}{2}=-\frac{1}{\left(-\frac{1}{\sqrt{3}}\right)}\left(x-\frac{1}{2}\right)$
$y=\sqrt{3} \cdot x-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}$

$y=\sqrt{3} \cdot x$

## Problem Kuznetsov Differentiation 17-12