## Precalculus (6th Edition) Blitzer

Given the rotation angle as $\theta=60^\circ$, we can obtain the rotation formulas as: $x=x'cos\theta -y' sin\theta = x'cos60^\circ -y' sin60^\circ =\frac{1}{2}(x'-\sqrt 3 y')$, $y=x'sin\theta +y' cos\theta = x'sin60^\circ -y' cos60^\circ =\frac{1}{2}(\sqrt 3x'+ y')$,