half-angle trig identity clarification I am working on the following trig half angle problem. I seem to be on the the right track except that my book answer shows -1/2 and I didn't get that in my answer. Where did I go wrong?
$$\sin{15^{\circ}} = $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \cos{30^{\circ}} }{ 2 } } $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \frac {\sqrt 3 }{ 2 } }{ 2 } } $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { \frac { 1 - \frac {\sqrt 3 }{ 2 } }{ 2 } (\frac {2} {2}) } $$
$$\sin \frac { 30^{\circ} }{ 2 } = \pm \sqrt { 2 - \sqrt {3} } $$
Book Answer
$$\sin \frac { 30^{\circ} }{ 2 } =  -\frac {1} {2}  \sqrt { 2 - \sqrt {3} } $$

$$\sqrt { \dfrac { 1 - \dfrac {\sqrt 3 }{ 2 } }{ 2 } \times \dfrac22} = \sqrt{\dfrac{2-\sqrt3}{\color{red}4}} = \dfrac{\sqrt{2-\sqrt3}}{\color{red}2}$$
