Show $1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$, where $z=\cos x+i \sin x$ Let $z=\cos x+i \sin x$. Show that 

$$1+z=2\cos \frac{1}{2} x(\cos \frac{1}{2}x +i \sin \frac{1}{2}x)$$

 A: Since
\begin{eqnarray}
\cos x&=&2\cos^2\frac{x}{2}-1\\
\sin x&=&2\sin\frac{x}{2}\cos\frac{x}{2},
\end{eqnarray}
we have
\begin{eqnarray}
1+z&=&1+\cos x+i\sin x\\
&=&1+\Big(2\cos^2\frac{x}{2}-1\Big)+i\Big(2\sin\frac{x}{2}\cos\frac{x}{2}\Big)\\
&=&2\cos^2\frac{x}{2}+2i\sin\frac{x}{2}\cos\frac{x}{2}\\
&=&2\cos\frac{x}{2}\Big(\cos\frac{x}{2}+i\sin\frac{x}{2}\Big)
\end{eqnarray}
A: That identity has a rather simple geometric interpretation. Consider the following diagram:

where $x$ is the red angle and
\begin{align}
z &= \cos x + i\cdot\sin x, \\
m &= \cos \frac{x}{2} + i\cdot\sin \frac{x}{2}, \\
n &= m \cdot \cos\frac{x}{2}.
\end{align}
In other words, $z$ is any point with $|z| = 1$, and $x$ is the angle it forms with line $01$. Point $m$ indicates the bisector of $x$ and $n$ is the orthogonal projection of $1$ onto line $0m$ (starting at $0$ we would need to go in the direction of $m$ for $\cos\frac{x}{2}$). That happens to be the orthogonal projection of $z$ onto $0m$ and thus also an intersection of segments $0m$ and $1z$.
We want to prove that $2\cdot n$ and $z+1$ represent the same point.
Observe, that because $|z|=1$, $\{0, 1, z, z+1\}$ constitute four corners of a rhombus in which bisectors are the diagonals, what implies that $0m$ actually passes through $z+1$.
Now we only need to prove that $|2\cdot n| = |z+1|$. But that is also true, because $n$ happens to be at the intersection of diagonals, which is also their midpoint.
I hope this helps $\ddot\smile$
