Make the content in the item visible when clicking the screen and left-align formula?

Question

I have two questions:

1) I can only let the items visible when clicking the screen, just adding up <2->. However, I don't know how the let the contents in the items do the same thing. It's not good to let them appear at the same time.

2) To make the formula be left. See my code, I have tried two methods but both of them fail.

More details, please see the picture and my code.

\documentclass[CJK]{beamer} 
\usepackage{beamerthemesplit} 
\usepackage[english]{babel}
\setbeamercovered{transparent}
\usepackage{amsmath}
\begin{document}
\begin{frame}
\frametitle{16.Given$f(x)=\sqrt{2}\cos (x-\frac{\pi}{12})$,$x\in \mathbb{R}$.\\
1)$f(-\frac{\pi}{6})=?$;
2)If $\cos \theta=\frac{3}{5}$,$\theta \in(\frac{3\pi}{2},2\pi)$, $f(2\theta+\frac{\pi}{3})$=?.}
%\vspace{-1cm}%delete the gap
\begin{itemize} \parsep 10pt %\itemsep 10pt
\item<2-> $1)f(-\frac{\pi}{6})=\sqrt{2}\cos (-\frac{\pi}{6}-\frac{\pi}  {12})=\sqrt{2}\cos (-\frac{\pi}{4})=\sqrt{2}\times \frac{\sqrt{2}}{2}=1.$\\
\item<3-> 2) ∵$\cos \theta=\frac{3}{5},\frac{3\pi}{2}<\theta<2\pi$,
∴$\sin \theta=-\sqrt{1-cos^2\theta}=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}$
∴ $\sin 2\theta=2\sin \theta \cos\theta=-\frac{24}{25}\Rightarrow \cos     2\theta=\cos^2\theta \sin^2\theta=-\frac{7}{25}$
 %\begin{flushleft} %fail to let the formula be left
 %f(2\theta+\frac{\pi}{3})&=\sqrt{2}\cos (2\theta+\frac{\pi}{3}-\frac{\pi}{12})\\
 %&=\sqrt{2}\cos (2\theta+\frac{\pi}{4})=\cos 2\theta-\sin2\theta\\
 %&=-\frac{7}{25}-(\frac{24}{25})=\frac{17}{25}.
 %\end{flushleft}
 \begin{flalign*} % fail again
 f(2\theta+\frac{\pi}{3})&=\sqrt{2}\cos (2\theta+\frac{\pi}{3}-\frac{\pi}{12})\\
 &=\sqrt{2}\cos (2\theta+\frac{\pi}{4})=\cos 2\theta-\sin2\theta\\
 &=-\frac{7}{25}-(\frac{24}{25})=\frac{17}{25}.
 \end{flalign*}
 \end{itemize}
 \end{frame}
 \end{document} 

Answer 1

Another method using the flalign environment would be to add && at the end of each line inside this environment. This combined with some \pause commands yields the following output:

And the MWE:

\documentclass[CJK]{beamer} 

\usepackage{beamerthemesplit} 
\usepackage[english]{babel}
\setbeamercovered{transparent}
\usepackage{amsmath}

\setbeamercovered{transparent=0}

\begin{document}

\begin{frame}
\frametitle{16.Given $f(x)=\sqrt{2}\cos (x-\frac{\pi}{12})$,$x\in \mathbb{R}$.\\
1) $f(-\frac{\pi}{6})=?$;
2) If $\cos \theta=\frac{3}{5}$,$\theta \in(\frac{3\pi}{2},2\pi)$, $f(2\theta+\frac{\pi}{3})$=?}

\pause1) $f(-\frac{\pi}{6})=\sqrt{2}\cos (-\frac{\pi}{6}-\frac{\pi}  {12})=\sqrt{2}\cos (-\frac{\pi}{4})=\sqrt{2}\times \frac{\sqrt{2}}{2}=1.$\\
\pause2) $\cos \theta=\frac{3}{5},\frac{3\pi}{2}<\theta<2\pi$
\pause$\sin \theta=-\sqrt{1-cos^2\theta}=-\sqrt{1-\frac{9}{25}}=-\frac{4}{5}$\\
\pause$\sin 2\theta=2\sin \theta \cos\theta=-\frac{24}{25}\Rightarrow \cos 2\theta=\cos^2\theta \sin^2\theta=-\frac{7}{25}$\pause
\begin{flalign}\nonumber
f(2\theta+\frac{\pi}{3})&=\sqrt{2}\cos (2\theta+\frac{\pi}{3}-\frac{\pi}{12})&&\\\nonumber
&=\sqrt{2}\cos (2\theta+\frac{\pi}{4})=\cos 2\theta-\sin2\theta&\\\nonumber
&=-\frac{7}{25}-(\frac{24}{25})=\frac{17}{25}.&&
\end{flalign}

\end{frame}
\end{document} 

Answer 2

regarding part 2 of your question, instead of the displayed flalign*, use ordinary in-line math (terminate the previous line with \\) and use aligned.

since aligned has a small extra space at the beginning (see Why is there a \, space at the beginning of the aligned environment?), begin the first line of the alignment with \! to negate it.

here's what that input would look like:

 \begin{itemize}
 ...
 \\
 $\begin{aligned}
  \! f(2\theta+\frac{\pi}{3})&=\sqrt{2}\cos (2\theta+\frac{\pi}{3}-\frac{\pi}{12})\\
  &=\sqrt{2}\cos (2\theta+\frac{\pi}{4})=\cos 2\theta-\sin2\theta\\
  &=-\frac{7}{25}-(\frac{24}{25})=\frac{17}{25}.
 \end{aligned}$
 \end{itemize}