## Algebra and Trigonometry 10th Edition

$sin~75°=\frac{\sqrt {2+\sqrt 3}}{2}$
$75°$ lies in the first quadrant. So, $sin~75°\gt0$: $sin~75°=sin\frac{150°}{2}=+\sqrt {\frac{1-cos~150°}{2}}=\sqrt {\frac{1-(-\frac{\sqrt 3}{2})}{2}}=\sqrt {\frac{2+\sqrt 3}{4}}=\frac{\sqrt {2+\sqrt 3}}{2}$