So I've got a trigonometry problem... maybe it's too easy but I can't understand it, can someone help me?
Suppose a regular pentagon of 5 sides is inside a circle of radius r
The pentagon is made of isosceles triangles. I should demonstrate that the area of the Pentagon is equal to $5r^2\cdot \sin (36^\circ) \cos (36^\circ)$
On
The formula is more general: for any regular n-gon inscribed in a circle of radius $r$, the area of the polygon is
$$r^2\sin\Bigl(\frac{180}n\Bigr)\cos\Bigl(\frac{180}n\Bigr) \quad\text{ (or }\;r^2\sin\Bigl(\frac{\pi}n\Bigr)\cos\Bigl(\frac{\pi}n\Bigr)\quad\text{if you work in radians}).$$
Indeed, the $n$-gon has area equal to $n$ times the area of the isosceles triangle with vertices the centre of the circle and two consecutive vertices of the $n$-gon. Denothing $s$ the length of a side of the $n$-gon and $a$ its apothem, the area of such a triangle is
$$\frac12 as=a\cdot\frac s2=r\cos\Bigl(\frac\pi n\Bigr)\cdot r\sin\Bigl(\frac\pi n\Bigr)=\frac12r^2\sin\Bigl(\frac{2\pi}n\Bigr).$$
By decomposing the pentagon into $5$ isosceles triangles and by writing the area of a triangle as $[ABC]=\frac{ab\sin C}{2}$ you have $[ABCDE]=\frac{5R^2}{2}\sin 72^\circ = 5R^2\sin36^\circ \cos 36^\circ$ as claimed (since $\sin(2x)=2\sin(x)\cos(x)$).