Theorem: Area of Sector

Let $\CC = ABC$ be a circle whose center is $A$ and with radii $AB$ and $AC$.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is $\theta$.
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Then the area $\AA$ of sector $BAC$ is given by:
:$\AA = \dfrac {r^2 \theta} 2$
where:
:$r = AB$ is the length of the radius of the circle
:$\theta$ is measured in radians.

Proof:
Let $\CC$ be a circle of radius $r$ whose center $A$ is at the origin and with radii $AB$ and $AC$.
Let $B$ and $C$ be arbitrary points on the circumference of $\CC$.
Let $C$ be positioned at $\tuple {0, r}$ so that $AC$ coincides with the $y$-axis.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is $\theta$.
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We have that $\CC$ is centered at the origin.
From Equation of Circle center Origin, the equation of $\CC$ is given by:
:$r^2 = x^2 + y^2$
Let $B$ have co-ordinates:
:$\tuple {x_0, y_0} = \tuple {x_0, \sqrt {r^2 - {x_0}^2} }$
Triangle $\triangle ADB$ is a right triangle whose leg $DB$ is opposite $\theta$, and whose hypotenuse is $AB$.
Hence:
{{begin-eqn}}
{{eqn | l = \sin \theta
      | r = \frac {\size {BD} } {\size {AB} }
      | c = {{Defof|Sine of Angle}}
}}
{{eqn | r = \frac {x_0} r
      | c = Definition of $x_0$ and $r$
}}
{{eqn | ll= \leadsto
      | l = \map \arcsin {\sin \theta}
      | r = \map \arcsin {\frac {x_0} r}
      | c = taking $\arcsin$ of both sides
}}
{{eqn | ll= \leadsto
      | l = \theta
      | r = \map \arcsin {\frac {x_0} r}
      | c = {{Defof|Real Arcsine}}
      | n = 1
}}
{{end-eqn}}
Consider the line $AB$, which by definition passes through $A = \tuple {0, 0}$ and $B = \tuple {x_0, \sqrt {r^2 - {x_0}^2} }$.
Let:
:$\Delta y$ be the change in $y$ between $A$ and $B$
:$\Delta x$ be the change in $x$
:$m$ be the slope of $AB$.
{{begin-eqn}}
{{eqn | l = \Delta y
      | r = \sqrt {r^2 - {x_0}^2} - 0
      | c = Difference between $y$ values of $A$ and $B$
}}
{{eqn | r = \sqrt {r^2 - {x_0}^2}
}}
{{eqn | l = \Delta x
      | r = x_0 - 0
      | c = Difference between $x$ values of $A$ and $B$
}}
{{eqn | r = x_0
}}
{{eqn | ll= \leadsto
      | l = m
      | r = \frac {\Delta y} {\Delta x}
      | c = {{Defof|Slope of Straight Line}}
}}
{{eqn | r = \frac {\sqrt {r^2 - {x_0}^2} } {x_0}
}}
{{eqn | l = y
      | r = \frac {\sqrt {r^2 - {x_0}^2} } {x_0} x
      | c = Equation of Straight Line in Plane in Slope-Intercept Form: $y$ intercept is $0$
      | n = 2
}}
{{end-eqn}}
Let $K$ be the region between $\CC$, $AC$ and $AB$, from $0$ to $x_0$, orange in the diagram above.
This is the sector of $\CC$ whose area we are to find.
Let $\AA$ be the area of $K$.
Then:
{{begin-eqn}}
{{eqn | l = \AA
      | r = \int_0^{x_0} \paren {\sqrt {r^2 - x^2} - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} x} \rd x
      | c = integrating between two curves to find area
}}
{{eqn | r = \int_0^{x_0} \sqrt {r^2 - x^2} \rd x - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \int_0^{x_0} x \rd x
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = \int_0^{x_0} \sqrt {r^2 - x^2} \rd x - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \sqbrk {\frac {x^2} 2}_0^{x_0}
      | c = Primitive of Power
}}
{{eqn | r = \sqbrk {\frac {x \sqrt {r^2 - x^2} } 2 + \frac {r^2} 2 \arcsin \frac x r}_0^{x_0} - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \sqbrk {\frac {x^2} 2}_0^{x_0}
      | c = Primitive of $\sqrt {a^2 - x^2}$
}}
{{eqn | r = \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2 + \frac {r^2} 2 \arcsin \frac {x_0} r - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \frac { {x_0}^2} 2
      | c = evaluating over range $\closedint 0 {x_0}$
}}
{{eqn | r = \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2 + \frac {r^2} 2 \arcsin \frac {x_0} r - \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2
      | c = simplifying
}}
{{eqn | r = \frac {r^2} 2 \arcsin \frac {x_0} r
      | c = cancelling
}}
{{eqn | r = \frac {r^2 \theta} 2
      | c = from $(1)$
}}
{{end-eqn}}
{{qed}}