<s>[INST] Maximum area of a isosceles triangle in a circle with a radius r As said in the title, I'm looking for the maximum area of a isosceles triangle in a circle with a radius $r$.
I've split the isosceles triangle in two, and I solve for the area $A=\frac{bh}{2}$*. I have made my base $x$, and solve for the height by using the Pythagorean theorem of the smaller triangle (seen in picture). 
$h=r+\sqrt{r^2-x^2}$

So my the formula, I think, for both triangles should be $A=x(r+\sqrt{r^2-x^2})$
But after I solved for the derivative, when put "$= 0$", and checked on my calculator, I got the maximum to be about $4.3301r$, which differs a lot from my book's answer of $\frac{3\sqrt{3}}{4}r^2$*. Is my formula for the area right? Am I going about this the wrong way, or is it just my derivative that is wrong? Thanks in advance 
*
 Edited from original post
$A=rx+x\sqrt{r^2-x^2}$
$A'=r+x(\frac{1}{2})(r^2-x^2)^{-\frac{1}{2}}(-2x)+\sqrt{r^2-x^2}$
$r+\sqrt{r^2-x^2}=\frac{x^2}{\sqrt{r^2-x^2}}$
$r\sqrt{r^2-x^2}+(r^2-x^2)=x^2$
$r^2+r\sqrt{r^2-x^2}=2x^2$
$r^2(r^2-x^2)=(2x^2-r^2)^2$
$r^4-r^2x^2=4x^2-4x^2r^2+r^4$
$4x^4=3x^2r^2$
$x=\frac{\sqrt{3}}{2}r$
 [/INST] 
We can also split the triangle into three smaller triangles using $\frac{1}{2} ab \sin C$. Therefore $\Delta ABC$ equals:
$$\frac{1}{2} r^2 \big(\sin \theta + \sin \theta + \sin(360º - 2 \theta ) \big)$$
$$\frac{1}{2} r^2 \big(2 \sin \theta + \sin(- 2 \theta ) \big)$$
Since the area is only dependent on $\theta$, we can take the derivative of:
$$f(\theta) = 2 \sin \theta + \sin(-2 \theta) \implies f' (\theta) = 2 \cos \theta - 2 \cos(-2\theta)$$
and setting $f'(\theta)$ equal to $0$:
$$\cos \theta - \cos(-2 \theta) = 0$$
$$\Rightarrow \cos \theta - (2 \cos^2 \theta - 1) = 0$$
$$\Rightarrow 2 \cos^2 \theta - \cos \theta - 1 = 0$$
$$\Rightarrow \cos \theta = -\frac{1}{2}, \cos \theta = 1$$
and the only solution that makes sense in the range $\theta = (0º, 180º)$ is $\theta = 120º$.
Therefore the maximum possible area is:
$$\frac{1}{2} r^2 \big(2 \sin 120º + \sin(- 2 \times 120º ) \big) = \frac{3 \sqrt3}{4} r^2.$$
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