Problem source

\begin{questionparts}
\item  Show that  $ 2\sin(\frac12\theta)=\sin \theta$ if and only if $\sin(\frac12\theta)=0\,$. 
\item Solve the equation $2\tan (\frac12\theta) = \tan\theta\,$.
 
\item Show that $2\cos(\frac12\theta)=\cos \theta$ if and only if $\theta=(4n+2)\pi\pm 2\phi$ where $\phi$ is 
defined by $\cos \phi=\frac12(\sqrt 3-1)\;$, $0\le \phi\le \frac{1}{2}\pi$, and $n$ is any integer. 
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$
\begin{align*}
&& 2 \sin (\tfrac12 \theta) &= \sin \theta \\
\Leftrightarrow && 2 \sin (\tfrac12 \theta) &= 2\sin (\tfrac12 \theta) \cos (\tfrac12 \theta) \\
\Leftrightarrow && 0 &= 2\sin(\tfrac12\theta)(1-\cos(\tfrac12 \theta)) \\
\Leftrightarrow && 0 = \sin(\tfrac12 \theta) &\text{ or } 1 = \cos(\tfrac12 \theta) \\
\Leftrightarrow && 0 &= \sin(\tfrac12 \theta)
\end{align*}

\item Let $= \tan(\tfrac12 \theta)$, then

\begin{align*}
&& 2t &= \frac{2t}{1-t^2} \\
\Leftrightarrow && 0 &= \frac{2t(1-(1-t^2)}{1-t^2} \\
&&&= \frac{2t^3}{1-t^2} \\
\Leftrightarrow && t&= 0 \\
\Leftrightarrow && \frac12\theta &= n \pi \\
\Leftrightarrow && \theta &= 2n\pi
\end{align*}

\item Let $c = \cos(\tfrac12 \theta)$, then

\begin{align*}
&& 2c &= 2c^2 - 1 \\
&& 0 &= 2c^2-2c-1 \\
\Leftrightarrow && c &= \frac{2 \pm \sqrt{4+8}}{4} \\
&&&= \frac{1 \pm \sqrt{3}}{2} \\
\Leftrightarrow && c &= \frac{1 - \sqrt{3}}{2} \\
\Leftrightarrow && \frac12 \theta &= \pm \cos^{-1} \frac{1 - \sqrt{3}}{2} + 2n \pi \\
&&&= \mp (\phi+\pi) + 2n \pi \\
\Leftrightarrow && \theta &= (4n+2)\pi \pm 2\phi

\end{align*}
\end{questionparts}