Ecuatia tangentei in punctul [tex]M(x_{0};y_{0})[/tex] este:
[tex]y-y_{0}=f'(x_0{})(x-x_{0})\\
y_{0}=f(\frac{\pi}{6})=\frac{1}{2}sin^2(\frac{\pi}{3})=\frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}\\
f'(x)=\frac{1}{2}\cdot 2sin(4x-\frac{\pi}{3})\cdot cos(4x-\frac{\pi}{3}) \cdot 4\\
f'(\frac{\pi}{3})=4\cdot sin\frac{\pi}{3}\cdot cos \frac{\pi}{3}=\sqrt{3}\\
y-\frac{3}{8}=\sqrt{3}(x-\frac{\pi}{6})\\
y=\sqrt{3}x-\frac{\pi\sqrt{3}}{6}+\frac{3}{8}[/tex]
