<s>[INST] 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
$$\sin \frac { 30^{\circ} }{ 2 } =  -\frac {1} {2}  \sqrt { 2 - \sqrt {3} } $$
 [/INST] In the $4^{th}$ equation, you should have $$\pm\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2} \cdot \frac{2}{2}}=\pm\sqrt{\frac{2-\sqrt{3}}{4}}=\pm\frac{1}{2}\sqrt{2-\sqrt{3}}.$$ Since $30/2$ is in the first quadrant, the answer should be the positive.
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