complex to standard form

Hi I was looking for steps to show z=sin(i ln(2)) + cos(i ln(2)) = 5/4 +3i/4

$\boxed{~ \begin{array}{rcll} \sin{z} &=& \frac{1}{2i} \cdot \left( e^{ix} - e^{-ix} \right) \\ \cos{z} &=& \frac{1}{2} \cdot \left( e^{ix} + e^{-ix} \right) \\ \end{array} ~}$

$\begin{array}{rcll} \sin{(i\cdot ln(2))} &=& \frac{1}{2i} \cdot \left( e^{i\cdot i\cdot ln(2)} - e^{-i\cdot i\cdot ln(2)} \right) \\ &=& \frac{1}{2i} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right) \qquad i\cdot i = i^2 = -1\\ \cos{(i\cdot ln(2))} &=& \frac{1}{2} \cdot \left( e^{i\cdot i\cdot ln(2)} + e^{-i\cdot i\cdot ln(2)} \right) \\ &=& \frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right) \qquad i\cdot i = i^2 = -1\\ \\ \sin{(i\cdot ln(2))}+ \cos{(i\cdot ln(2))}&=& \frac{1}{2i} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right) +\frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right)\\ &=& \frac{1}{2i} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right) \cdot \frac{i}{i} +\frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right)\\ &=& \frac{i}{-2} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right) +\frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right)\\ &=& \frac{1}{2} \cdot \left( e^{- ln(2)} + e^{ ln(2)} \right) - \frac{i}{2} \cdot \left( e^{- ln(2)} - e^{ ln(2)} \right)\\\\ e^{ ln(2)} = 2 && e^{- ln(2)} = \frac{1}{e^{ ln(2)}} = \frac12\\\\ \sin{(i\cdot ln(2))}+ \cos{(i\cdot ln(2))} &=&\frac{1}{2} \cdot \left( \frac12 + 2 \right) - \frac{i}{2} \cdot \left( \frac12 - 2 \right)\\ &=&\frac{5}{4} - \frac{i}{2} \cdot \left( -\frac32 \right) \\ &=&\frac{5}{4} + i \cdot \left( \frac34 \right) \end{array}$

as a guest I couldn't rate it .... so thanks!!!