4
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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) $\endgroup$ – Winther May 14 '18 at 14:19
  • $\begingroup$ Hint: $sin(x) = -i \frac{(e^{ix} - e^{-ix})}{2}$ $\endgroup$ – Fibonacci May 14 '18 at 14:22
2
$\begingroup$

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}

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Ennar May 14 '18 at 14:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.