Max perimeter of triangle inscribed in a circle

What is the maximum perimeter of a triangle inscibed in a circle of radius $1$?

I can’t seem to find a proper equation to calculate the derivative.

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Use law of sines and this

Intuitively, the maximum ought to be an equilateral triangle, with perimeter $3\sqrt 3$.

If you want to use calculus, let $\theta$ and $\phi$ be the arcs spanned by two of the sides, and calculate the perimeter as
$$ 2\sin \frac{\theta}2 + 2\sin\frac\phi2 + 2\sin\frac{2\pi-\theta-\phi}2 $$
Some manipulation of trigonometric identities will be involved.

Hint: the triangle can be defined by two numbers $0 < \theta < \phi < 2 \pi$, where one vertex $A$ say is at (1,0) and vertices $B, C$ are at $(\cos \theta, \sin \theta) \mbox{ and } (\cos \phi, \sin \phi)$.

Use Pythagoras to calculate AB + BC + CA, and differentiate with respect to $\theta, \phi$ to get maximum.