Originally Posted by

**Richmond** Show that

part a)

sin(x+iy) = sin x cosh y+ i cos x sinh y

using complex exponentials

This is what i got so far

LHS

sin(x+iy)

$\displaystyle

\frac{e^{ix-y} - e^{-ix+y}}{2i}

$

RHS

sin x cosh y + i cos x sinh y

$\displaystyle

\frac{e^{ix+y} + e^{ix-y} - e^{-ix+y} - e^{-ix-y}}{4i} + i\frac{e^{ix+y} - e^{ix-y} + e^{-ix+y} - e^{-ix-y}}{4}

$

when i combine(cross multiply) both the RHS equation i get

$\displaystyle

\frac{8e^{ix-y} + 8e^{-ix+y}}{16i}

$

Which is equal to

$\displaystyle

\frac{e^{ix-y} - e^{-ix+y}}{2i}

$

LHS = RHS

part b)

If z = x+iy with x and y real and if sin(z) = 2, deduce that cos x = 0

how do I go upon that?