Showing $r^2\left(\frac\theta2-\sin\frac\theta2\cos\frac\theta2\right)=\frac{r^2}{2}(\theta-\sin\theta)$

Question

I derived the formula for the area of a circular segment as follows:

However, the formula I find online when using radians is: A = $\frac{r^2}{2}\bigl(\theta - \sin(\theta)\bigr)$. Is there a simpler way of deriving the formula or can the first formula (using radians) be converted into the second?

Answer 1

From $A = r^2(\frac{\theta}{2} - \sin(\frac{\theta}{2})\cos(\frac{\theta}{2}))$, remember that $2\sin\theta\cos\theta = \sin(2\theta)$.

Rearranging it and substituting $\theta=\frac{\theta}{2}$, we get: $\sin(\frac{\theta}{2})\cos(\frac{\theta}{2}) = \frac{1}{2}\sin\theta$

Putting that into our area of a circular segment formula (radians), we get:

$A = r^2(\frac{\theta}{2} - \frac{1}{2}\sin\theta)$

Factoring out $\frac{1}{2}$, we get

$\fbox{$A=\frac{r^2}{2}(\theta - \sin\theta)$}$