## Algebra and Trigonometry 10th Edition

$sin~67°~30'=\frac{\sqrt {2+\sqrt 2}}{2}$ $cos~67°~30'=\frac{\sqrt {2-\sqrt 2}}{2}$ $tan~67°~30'=\sqrt 2+1$
$67°~30'$ lies in the first quadrant. $sin~67°~30'=sin\frac{135°}{2}=+\sqrt {\frac{1-cos~135°}{2}}=\sqrt {\frac{1-(-\frac{\sqrt 2}{2})}{2}}=\sqrt {\frac{2+\sqrt 2}{4}}=\frac{\sqrt {2+\sqrt 2}}{2}$ $cos~67°~30'=cos\frac{135°}{2}=+\sqrt {\frac{1+cos~135°}{2}}=\sqrt {\frac{1-\frac{\sqrt 2}{2}}{2}}=\sqrt {\frac{2-\sqrt 2}{4}}=\frac{\sqrt {2-\sqrt 2}}{2}$ $tan~67°~30'=tan\frac{135°}{2}=\frac{1-cos~135°}{sin~135°}=\frac{1-(-\frac{\sqrt 2}{2})}{\frac{\sqrt 2}{2}}=\frac{2+\sqrt 2}{\sqrt 2}=\sqrt 2+1$