Complex Numbers

Could someone explain the very last step of these proofs;

ie. How:

\frac{i^{n-k} - (-i)^{n-k}}{2i} = sin(\frac{1}{2}(n-k)\pi)

and

\frac{i^{n-k} + (-i)^{n-k}}{2} = cos(\frac{1}{2}(n-k)\pi)

Could someone explain the very last step of these proofs;

ie. How:

\frac{i^{n-k} - (-i)^{n-k}}{2i} = sin(\frac{1}{2}(n-k)\pi)

and

\frac{i^{n-k} + (-i)^{n-k}}{2} = cos(\frac{1}{2}(n-k)\pi)

\displaystyle i=e^{i\frac{\pi}{2}}

\displaystyle -i=e^{-i\frac{\pi}{2}}

and

\displaystyle cosh(ix)=cosx

\displaystyle \frac{1}{i}sinh(ix)=sinx

\displaystyle i=e^{i\frac{\pi}{2}}

\displaystyle -i=e^{-i\frac{\pi}{2}}

and

\displaystyle cosh(ix)=cosx

\displaystyle \frac{1}{i}sinh(ix)=sinx

Thank you, I haven't covered hyperbolic functions yet but was able to understand this with a quick peak at the chapter.

Look back to the first line i.e.

\sin(mx) = \frac{e^{i m x} - e^{-imx}}{2i}

and put

m=\frac{1}{2}(n-k)

and

x=\pi

to get

\sin(\frac{1}{2}(n-k)\pi) = \frac{(e^{i \pi})^{\frac{1}{2}(n-k)} - (e^{i\pi})^{\frac{-1}{2}(n-k)}}{2i}

Then use

e^{i \pi} = -1

to get

\frac{((e^{i \pi})^{\frac{1}{2}})^{(n-k)} - ((e^{i\pi})^{\frac{-1}{2}})^{(n-k)}}{2i} = \frac{((-1)^{\frac{1}{2}})^{(n-k)} - ((-1)^{\frac{-1}{2}})^{(n-k)}}{2i} = \frac{i^{(n-k)} - (i^{-1})^{(n-k)}}{2i} = \frac{i^{(n-k)} - (-i)^{(n-k)}}{2i}

Similarly for the other one.

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