Proving that $\sin(z+w) = \sin(w) \cos(z) + \sin(z) \cos(w)$ using complex exponentials

Let's define:

$$\sin(z) = \frac{\exp(iz) - \exp(-iz)}{2i}$$ $$\cos(z) = \frac{\exp(iz) + \exp(-iz)}{2}$$

We are to prove that $$\sin(z+w)=\sin(w) \cos(z) + \sin(z)\cos(w), \forall_{z,w \in \mathbb{C}}$$ using only the following statement: $\exp(z+w) = \exp(w)\exp(z)$.

I managed only to show that: $$\sin(z + w) = \frac{\exp(iz)\exp(iw)}{2i} - \frac{\exp(-iz)\exp(-iw)}{2i}.$$
Where can I go from here?

• Compute the right hand side $\sin(w)\cos(z) + \sin(z)\cos(w)$ and compare! (small typo: you are missing an $i$ in the denominator of your $\sin(z+w)$ expression) – Winther May 14 '18 at 14:19
• Hint: $sin(x) = -i \frac{(e^{ix} - e^{-ix})}{2}$ – Fibonacci May 14 '18 at 14:22

Since OP starts from left-hand-side and asks "where can I go from here", I'll give a solution staring from the left-hand-side.

\begin{align} & \sin(z+w) \\ &= \frac{\exp(i(z+w)) - \exp(-i(z+w))}{2i} \\ &= \frac{\exp(iz)\exp(iw) - \exp(-iz)\exp(-iw)}{2i} \\ &= \frac{\exp(iz)\exp(iw) \color{blue}{-\exp(iz)\exp(-iw) + \exp(iz)\exp(-iw)} - \exp(-iz)\exp(-iw)}{4i} \\ &+ \frac{\exp(iz)\exp(iw) \color{blue}{-\exp(-iz)\exp(iw) + \exp(-iz)\exp(iw)} - \exp(-iz)\exp(-iw)}{4i} \\ &= \frac{\exp(iz)(\exp(iw)-\color{blue}{\exp(-iw)}) + \exp(-iz)(\color{blue}{\exp(iw)}-\exp(-iw))}{4i} \tag{terms 1,2,7,8} \\ &+ \frac{\exp(-iw)(\color{blue}{\exp(iz)}-\exp(-iz)) + \exp(iw)(\exp(iz) - \color{blue}{\exp(-iz)})}{4i} \tag{terms 3-6} \\ &= \frac{(\exp(iz)+\exp(-iz))(\exp(iw)-\color{blue}{\exp(-iw)})}{2 \cdot 2i} \\ &+ \frac{(\exp(iw)+\exp(-iw))(\color{blue}{\exp(iz)}-\exp(-iz))}{2 \cdot 2i} \\ &= \cos(z)\sin(w) + \cos(w)\sin(z) \end{align}

Consider the RHS,

$$\sin z\cos w+\cos z\sin w$$

$$=\frac{e^{iz}-e^{-iz}}{2i}\frac{e^{iw}+e^{-iw}}{2}+\frac{e^{iw}-e^{-iw}}{2i}\frac{e^{iz}+e^{-iz}}{2}$$

[By doing the usual calculation which is skipped]

$$=2\frac{e^{i(z+w)}-e^{-i(z+w)}}{4i}$$ $$=\frac{e^{i(z+w)}-e^{-i(z+w)}}{2i}$$ $$=\sin(z+w)$$

Proved.

It is easier to start from $\sin z\cos w + \cos z\sin w$, but you can also directly continue from what you have.

It is straightforward to check $e^{iz} = \cos z + i\sin z$ directly from the definition of $\cos$ and $\sin$. Also, quite easily seen, $\cos$ is even, while $\sin$ is odd.

So, from what you have:

\begin{align}\sin(z + w) &= \frac{\exp(iz)\exp(iw)}{2i} - \frac{\exp(-iz)\exp(-iw)}{2i} \\ &= \frac 1{2i}( (\cos z + i\sin z)(\cos w + i\sin w) - (\cos z - i\sin z)(\cos w - i\sin w) )\\ &= \sin z\cos w + \cos z\sin w.\end{align}

• As a bonus, you can see how terms from $\cos(z+w) = \cos z\cos w - \sin z\sin w$ cancel, while if you removed $i$ from denominator and changed $-$ to $+$ in the first line, it would be the other way around, terms from $\sin(z+w)$ would cancel, giving you the addition formula for cosine. – Ennar May 14 '18 at 14:54