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Note that \begin{aligned} \sin(z-w) &= \frac{e^{i(z-w)}-e^{-i(z-w)}}{2i} \\ &= \frac{e^{iz}e^{-iw}-e^{-iz}e^{iw}+e^{iz}e^{iw}-e^{iz}e^{iw}}{2i} \\ &= \frac{e^{iw}(e^{iz}-e^{-iz})}{2i}-\frac{e^{iz}(e^{iw}-e^{-iw})}{2i} \\ &= e^{iw}\sin(z)-e^{iz}\sin(w) \\ &= 2(\cos(w) \sin(z)-\cos(z) \sin(w))+e^{-iz}\sin(w)-e^{iw} \sin(z) \end{aligned}

I have omitted to add $e^{-iw} \sin(z)-e^{-iw} \sin (z)$

The idea was to achieve that $e^{-iz} \sin(w)-e^{iw} \sin (z) = -\sin (z-w)$ to clear this value and easily obtain the equality, but using the sin parity is not given, any suggestions?

Wrloord
  • 1,626

4 Answers4

3

Note that we have

$$\sin(x)=\text{Im}(e^{ix})$$

Then, we can write

$$\begin{align} \sin(z-w)&=\text{Im}(e^{i(z-w)})\\\\ &=\text{Im}\left(e^{iz}e^{-iw}\right)\\\\ &=\sin(z)\cos(w)-\sin(w)\cos(z) \end{align}$$

as was to be shown!

Mark Viola
  • 179,405
2

Just expand $e^{iw}$ and $e^{iz}$ and the imaginary terms will cancel out.

$\begin{aligned} \sin(z-w) &= \frac{e^{i(z-w)}-e^{-i(z-w)}}{2i} \\ &=...\\ &= e^{iw}\sin(z)-e^{iz}\sin(w) \\ &= (\cos(w)+i\sin(w))\sin(z)-(\cos(z)+i\sin(z))\sin(w) \\ &= \cos(w)\sin(z)-\cos(z)\sin(w)+i(\sin(w)\sin(z)-\sin(z)\sin(w)) \\ &= \cos(w)\sin(z)-\cos(z)\sin(w) \\ \end{aligned} $

marty cohen
  • 107,799
1

You also have \begin{aligned} \sin(z-w) &= \frac{e^{i(z-w)}-e^{-i(z-w)}}{2i} \\ &= \frac{e^{iz}e^{-iw} - e^{-iz}e^{iw} + e^{-iz}e^{-iw} - e^{-iz}e^{-iw}}{2i} \\ &= \frac{e^{-iw}(e^{iz} - e^{-iz})}{2i} - \frac{e^{-iz}(e^{iw} - e^{-iw})}{2i} \\ &= e^{-iw}\sin(z)-e^{-iz}\sin(w). \end{aligned}

Therefore \begin{aligned} \sin(z-w) &= \frac12\sin(z-w) + \frac12\sin(z-w) \\ &= \frac12\left(e^{iw}\sin(z)-e^{iz}\sin(w)\right) + \frac12\left(e^{-iw}\sin(z) - e^{-iz}\sin(w)\right) \\ &= \frac{e^{iw} + e^{-iw}}{2}\sin(z) - \frac{e^{iz} + e^{-iz}}{2}\sin(w) \\ &= \cos(w)\sin(z) - \cos(z)\sin(w). \end{aligned}

But the other answers seem like less work.

Less symmetric, but quicker, you already have

\begin{aligned} \sin(z-w) &= e^{iw}\sin(z)-e^{iz}\sin(w) \\ &= 2(\cos(w) \sin(z)-\cos(z) \sin(w)) + \underbrace{e^{-iz}\sin(w)-e^{iw} \sin(z)}_{-\sin(z - w)}. \end{aligned}

Add $\sin(z - w)$ to both sides:

$$ 2\sin(z - w) = 2(\cos(w) \sin(z)-\cos(z) \sin(w)), $$

then divide by $2$.

David K
  • 98,388
0

After your first formula block, you could just take the real part, as the left side is a real function.

Or you could resolve the exponentials into the Euler formula $e^{iw}=\cos w+i\sin w$ and then recognize that the resulting two $i\sin w\sin z$ terms cancel. You went one step too far into this direction.

Lutz Lehmann
  • 126,666