Proving a Complex Identity: $ e^{i\theta} + e^{i\phi} = 2\cos\left[\frac{\theta-\phi}{2}\right]e^{\frac{i(\theta+\phi)}{2}} $

Question

I've been attempting the following problem for quite a while to no avail.

Prove that $$ e^{i\theta} + e^{i\phi} = 2\cos\left[\frac{\theta-\phi}{2}\right]e^{\frac{i(\theta+\phi)}{2}} $$ (i) by calculation, and (ii) geometrically.

(ii) was easy for me as $e^{i\theta}$ and $e^{i\phi}$ are both on the unit circle, have the same magnitude, the interior angle of the "parallelogram formed by addition" is $\theta-\phi$. My question is with (i). I wrote

$$ \cos \theta + i\sin \theta + \cos \phi + i\sin \phi = 2\cos\left[\frac{\theta-\phi}{2}\right] \left(\cos \frac{\theta+\phi}{2} + i\sin \frac{\theta+\phi}{2} \right) .$$ From here I'm not sure to use a half angle property or something else. How should I continue from here?

Answer 1

$$\Large\begin{align}2\cos \left[ \frac { \theta -\phi }{ 2 } \right] e^{ \frac { i(\theta +\phi ) }{ 2 } } & =2\cdot \frac { { e }^{ \frac { \theta -\phi }{ 2 } i }+{ e }^{ -\frac { \theta -\phi }{ 2 } i } }{ 2 } \cdot e^{ \frac { i(\theta +\phi ) }{ 2 } }\\ &={ e }^{ \frac { i\theta -i\phi +i\theta +i\phi }{ 2 } }+{ e }^{ \frac { -i\theta +i\phi +i\theta +i\phi }{ 2 } } \\ &={ e }^{ i\theta }+{ e }^{ i\phi }\end{align}$$

Answer 2

Now, all you have to do is\begin{align*}\cos\theta+\cos\phi&=\cos\left(\frac{\theta+\phi}2+\frac{\theta-\phi}2\right)+\cos\left(\frac{\theta+\phi}2-\frac{\theta-\phi}2\right)\\&=2\cos\left(\frac{\theta+\phi}2\right)\cos\left(\frac{\theta-\phi}2\right)\end{align*}and\begin{align*}\sin\theta+\sin\phi&=\sin\left(\frac{\theta+\phi}2+\frac{\theta-\phi}2\right)+\sin\left(\frac{\theta+\phi}2-\frac{\theta-\phi}2\right)\\&=2\sin\left(\frac{\theta+\phi}2\right)\cos\left(\frac{\theta-\phi}2\right).\end{align*}

Answer 3

Let $\phi> \theta$, from drawing a picture, then obtuse angle between is $\pi - (\phi - \theta) $ , to find magnitude of the resultant vector, we use law of cosines:

$$ c^2 = a^2 + b^2 -2ab \cos \theta= 1 + 1 - 2 \cos \left(\pi -( \phi - \theta) \right) = 2 + 2 \cos( \phi - \theta)$$

We find $c= \sqrt{2} \sqrt{ 1 + \cos ( \phi - \theta ) }= 2\cos \frac{\phi- \theta}{2}$, now we need the angle between the vectors which turns out to be $\tan^{-1} \frac{\sin (\phi ) + \sin \theta}{\cos \phi + \cos \theta} = \frac{\phi+\theta}{2}$, hence the final answer is:

$$ e^{ i \theta} + e^{ i \phi} = 2 \cos \frac{\phi - \theta}{2} e^{i \left( \frac{ \phi + \theta}{2} \right)}$$