## Algebra and Trigonometry 10th Edition

$x=2r(1-cos~θ)$
Power Reducing Formula (page 543): $sin^2x=\frac{1-cos~2x}{2}$ Now, make $x=\frac{θ}{2}$: $sin^2\frac{θ}{2}=\frac{1-cos~θ}{2}$ $\frac{x}{2}=2r~sin^2\frac{θ}{2}$ $\frac{x}{2}=2r~\frac{1-cos~θ}{2}$ $\frac{x}{2}=r(1-cos~θ)$ $x=2r(1-cos~θ)$