Find the biggest area of an isosceles triangle in an unit circle An isosceles triangle is inscribed in the unit circle. Determine the largest area of ​​the triangle can assume.
My solution attempt:
An isosceles triangle has two sides of equal size, and the sides will be at some point on the circumference of the unit circle during suggestion the $x$-axis. The height will be $1 + h \le 2$ and width $b + 1 \le 2$
Another way would be to make the vectors of the three points that will be on the periphery (assuming that it can provide the largest area on the periphery)
I really do not know how to solve it.
 A: Hint
The triangle is $\Delta ABC$: $AB=AC=b$ and $BC=a$. 
Also $\angle A=2x \rightarrow \angle B=\frac{\pi - 2x}{2}=\frac{\pi}{2}-x$, so:
$$S(ABC)=\frac{b^2\sin 2x}{2}$$
And by sine rule:
$$\frac{b}{\sin \angle B}=2\cdot 1 \rightarrow b=2\cdot \cos x$$ 
and then:
$$S(ABC)=2\cdot \cos^2x\cdot\sin 2x=(1+\cos 2x)\cdot \sin 2x$$
Can you finish?
A: equilateral triangle has maximum area
side would be $\sqrt{3}$ by using trigonometry
area is $\frac{\sqrt 3}{4}a^2$$=$$\frac{3\sqrt 3}{4}$
A: I assumed a central angle within the circle A to create an additional isosceles triangle and find the base N, as well as find the total height. 
Because this is a unit circle, the radius is 1.
With the central angle A, we create the isosceles triangle with base N and with base angles X and with height K.
Drawing a perpendicular bisector down from the vertex of A creates two right triangles; each has height K, base angle X, upper angle (for lack of a better term) a/2, and hypotenuse of 1 (because the radius is 1). With this, we can create some functions/equations based on the variables:
X = 90 - (A/2)
sin(x) = K
cos(x) = n/2
2cos(x) = n
The total height of the isosceles triangle in question (the one with the largest area) is given by K+1.
The base of the isosceles triangle with base angle X is given by N (as is the largest possible isosceles triangle).
The area of a triangle is given by ((base*height)/2).
Therefore, we can perform substitutions, and get that
(2cos(x)(sin(x) + 1))/2 = Area.
More simply, cos(x)(sin(x) + 1) = Area.
However, we established these relations based on an assumed central angle; therefore, we need to substitute.
cos(90 - (a/2))(sin(90 - (a/2)) + 1) = Area.
Proceed from there.
A: First note that the two angles in the figure below must be equal (the two triangles in the lower part of the picture are congruent by side-angle-side). Denote their common value by $\theta$.
The area of the triangle is then $A(\theta) = \frac{\sin2\theta}{2}+\sin\theta$, where  $0<\theta<\pi$.
Now find the maximum of $A$ on its domain.

