## Problem Statement

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

\[
\left\{\begin{array}{l}
x=2 \operatorname{tg} t \\
y=2 \sin ^{2} t+\sin 2 t
\end{array}\right.
\]

$t_{0}=\frac{\pi}{4}$
## Solution

$x_{0}=x\left(t_{0}\right)=2 \operatorname{tg} \frac{\pi}{4}=2$

$y_{0}=y\left(t_{0}\right)=2 \sin ^{2} \frac{\pi}{4}+\sin \frac{\pi}{2}=2$

$y_{x}^{\prime}=\frac{y^{\prime}(t)}{x^{\prime}(t)}=\frac{4 \sin t \cdot \cos t+2 \cos 2 t}{\frac{2}{\cos ^{2} t}}=\frac{4 \sin t \cdot \cos ^{3} t+2 \cos 2 t \cdot \cos ^{2} t}{2}$

$y_{0}^{\prime}=y^{\prime}\left(t_{0}\right)=\frac{4 \sin \frac{\pi}{4} \cdot \cos ^{3} \frac{\pi}{4}+2 \cos \frac{\pi}{2} \cdot \cos ^{2} \frac{\pi}{4}}{2}=\frac{1}{2} \cdot 4 \cdot \frac{1}{\sqrt{2}} \cdot\left(\frac{1}{\sqrt{2}}\right)^{3}=\frac{1}{2}$

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

- Equation of the normal.

$y-2=\frac{-1}{\frac{1}{2}}(x-2) \Longleftrightarrow y=-2 x+4+2 \Longleftrightarrow y=-2 x+6$

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

$y-2=\frac{1}{2}(x-2) \Longleftrightarrow y=\frac{1}{2} x-1+2 \Longleftrightarrow y=\frac{1}{2} x+1$