Question

$\red{ \rm{sin(x) \left( \dfrac{ {e}^{ - y } + {e}^{y} }{2} \right) + \cos(x) \left( \dfrac{ {e}^{ - y} - {e}^{y} }{2i} \right)}}$​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{sin(x)\left(\dfrac{e^{-y}+e^{y}}{2}\right)+cos(x)\left(\dfrac{e^{-y}-e^{y}}{2i}\right)}$

$\rm{=\dfrac{e^{ix}-e^{-ix}}{2i}\cdot\dfrac{e^{-y}+e^{y}}{2}+\dfrac{e^{ix}+e^{-ix}}{2}\cdot\dfrac{e^{-y}-e^{y}}{2i}}$

$\rm{=\dfrac{\left(e^{ix}-e^{-ix}\right)\left(e^{-y}+e^{y}\right)}{4i}+\dfrac{\left(e^{ix}+e^{-ix}\right)\left(e^{-y}-e^{y}\right)}{4i}}$

$\rm{=\dfrac{e^{-y+ix}-e^{-y-ix}+e^{y+ix}-e^{y-ix}}{4i}+\dfrac{e^{-y+ix}+e^{-y-ix}-e^{y+ix}-e^{y-ix}}{4i}}$

$\rm{=\dfrac{e^{-y+ix}-e^{-y-ix}+e^{y+ix}-e^{y-ix}+e^{-y+ix}+e^{-y-ix}-e^{y+ix}-e^{y-ix}}{4i}}$

$\rm{=\dfrac{2\,e^{-y+ix}-2\,e^{y-ix}}{4i}}$

$\rm{=\dfrac{e^{-y+ix}-e^{y-ix}}{2i}}$

$\rm{=-\,\dfrac{e^{y-ix}-e^{-y+ix}}{2i}}$

$\rm{=-\,\dfrac{e^{y-ix}-e^{-(y-ix)}}{2i}}$

$\rm{=-\,sinh\left(y-ix\right)}$