Answer
$sin~112°~30'=\frac{\sqrt {2+\sqrt 2}}{2}$
$cos~112°~30'=-\frac{\sqrt {2-\sqrt 2}}{2}$
$tan~112°~30'=-(1+\sqrt 2)$
Work Step by Step
$112°~30'$ lies in the second quadrant.
$sin~112°~30'=sin\frac{225°}{2}=+\sqrt {\frac{1-cos~225°}{2}}=\sqrt {\frac{1-(-\frac{\sqrt 2}{2})}{2}}=\sqrt {\frac{2+\sqrt 2}{4}}=\frac{\sqrt {2+\sqrt 2}}{2}$
$cos~112°~30'=cos\frac{225°}{2}=-\sqrt {\frac{1+cos~225°}{2}}=-\sqrt {\frac{1-\frac{\sqrt 2}{2}}{2}}=-\sqrt {\frac{2-\sqrt 2}{4}}=-\frac{\sqrt {2-\sqrt 2}}{2}$
$tan~112°~30'=tan\frac{225°}{2}=\frac{1-cos~225°}{sin~225°}=\frac{1-(-\frac{\sqrt 2}{2})}{-\frac{\sqrt 2}{2}}=\frac{2+\sqrt 2}{-\sqrt 2}=-\sqrt 2-1=-(1+\sqrt 2)$
