# 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 $$\sin \frac { 30^{\circ} }{ 2 } = -\frac {1} {2} \sqrt { 2 - \sqrt {3} }$$

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As $\sin{30^{\circ}} = 1/2$, I do not know how $\sin{15^{\circ}}$ is to equal $-1/2$. You have the right idea. – Ron Gordon Feb 5 '13 at 2:20
You're close! When you multiplied by $(2/2)$, you should have gotten a denominator of $4$, which can be pulled out of the radical as a multiplied $1/2$. Since $15^\circ$ is in the first quadrant, the sign of the sine should be positive, so the "book answer" is wrong in that regard. – Blue Feb 5 '13 at 2:20
@Blue thanks for the clarification :) – sam Feb 5 '13 at 2:30

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