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Compute area A given that A is a triangle inscribed in a unit circle with one vertex at $P = (1,0)$ and the other two at the points $P_1$ and $P_2$ given by polar coordinates with angles $\theta_1$ and $\theta_2$, where $\theta_2 > \theta_1$.

My attempt at a solution:

I attempted to take the cross product of $\vec{PP_1}$ and $\vec{PP_2}$ by setting a $z$ coordinate for both vectors as $0$:

$\vec{B} =\vec{PP_1} \times \vec{PP_2} = (\sin(\theta_2)(1-\cos(\theta_1)) - \sin(\theta_1)(1-\cos(\theta_2)))\hat{k}$

$A = 0.5|\vec{B}| = 0.5(\sin(\theta_1 - \theta_2) + \sin(\theta_2) - \sin(\theta_1)$

Is there a more efficient way to do this problem? The polar coordinates could be converted to (x,y) to use Heron's Formula, but it does not seem to be the most optimal way.

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This seem pretty efficient to me. You can get the same formula directly by noticing that $$A_{\triangle POP_1}+A_{\triangle P_1OP_2}=A_{\triangle POP_2}+A_{\triangle PP_2P_1}$$ On the unit circle, these areas are $\frac{1}{2}\sin\theta_1$, $\frac{1}{2}\sin(\theta_2-\theta_1)$, $\frac{1}{2}\sin\theta_2$, and $A$

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