如何在博客园中插入公式

\begin{align*}\sin^2 A+\sin^2B+\sin^2C &=\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos 2C}{2}\\ &=\frac{3}{2}-\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C)\\ &=\frac{3}{2}-\cos(A+B)\cos(A-B)-\cos^2C+\frac{1}{2}\\ &=2+\cos C\cos(A-B)-\cos^2C \\ &\leq 2+|\cos C|-\cos^2C \\ &=-(|\cos C|-\frac{1}{2})^2+\frac{9}{4}\\ &\leq \frac{9}{4}.\end{align*}

 

\begin{align}a^2+b^2&=c^2\\\beta^2&=\sum_{n=0}^{\infty}{\frac{1}{n}}+\int_0^1{\frac{x}{e^x-1}dx}\end{align}

 {\sigma}_{k}(n)=\sum_{d|n}^{}{d}^{k}

\begin{align}
  (a + b)^3  &= (a + b) (a + b)^2        \\
             &= (a + b)(a^2 + 2ab + b^2) \\
             &= a^3 + 3a^2b + 3ab^2 + b^3
\end{align}

 \begin{flalign}
& q=
  \begin{cases}
    0 & \mbox{ if $(A+B)\bmod 2 =0$}\\  第一个分支
    1 & \mbox{ if $(A+B)\bmod 4 =0$};\\ 第二个分支
    -1 & \mbox{ if $(A-B)\bmod 2=0$}\\  第三个分支
  \end{cases}  & \nonumber
\end{flalign}

 \begin{align}

E(S^2) &=E\left(\frac{1}{n} \sum_{i=1}^n (X_i-\bar{X})^2\right) \\ & =E\left(\frac{1}{n}\sum_{i=1}^n X_i^2\right) - E\left(\frac{1}{n}\sum_{i=1}^n 2\bar{X}X_i\right) + E\left(\frac{1}{n}\sum_{i=1}^n \bar{X}^2\right) \\ & =EX^2 -E(\bar{X}^2) \\ & =DX + (EX)^2 - D\bar{X} - (E\bar{X})^2 \\ & =\frac{n-1}{n}DX \end{align}

\begin{equation*}
  \begin{rcase}
    B' &= -\partial\times E          \\
    E' &=  \partial\times B - 4\pi j \,
  \end{rcase}
  \quad \text {Maxwell's equations}
\end{equation*}

 

\begin{equation} \begin{aligned}
  V_j &= v_j                      &
  X_i &= x_i - q_i x_j            &
      &= u_j + \sum_{i\ne j} q_i \\
  V_i &= v_i - q_i v_j            &
  X_j &= x_j                      &
  U_i &= u_i
\end{aligned} \end{equation}

 

\begin{align*} f(x) & = (x+a)(x+b) \\ & = x^2 + (a+b)x + ab \end{align*}

\begin{gather*} E(X)=\lambda \qquad D(X)=\lambda \\ E(\bar{X})=\lambda \\ D(\bar{X})=\frac{\lambda}{n} \\ E(S^2)=\frac{n-1}{n}\lambda \\ \end{gather*}

\begin{align*}\sin^2 A+\sin^2B+\sin^2C &=\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos 2C}{2}\\ &=\frac{3}{2}-\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C)\\ &=\frac{3}{2}-\cos(A+B)\cos(A-B)-\cos^2C+\frac{1}{2}\\ &=2+\cos C\cos(A-B)-\cos^2C \\ &\leq 2+|\cos C|-\cos^2C \\ &=-(|\cos C|-\frac{1}{2})^2+\frac{9}{4}\\ &\leq \frac{9}{4}.\end{align*}

 

\begin{align}a^2+b^2&=c^2\\\beta^2&=\sum_{n=0}^{\infty}{\frac{1}{n}}+\int_0^1{\frac{x}{e^x-1}dx}\end{align}

 

参考链接如下：
http://www.cnblogs.com/cmt/p/3279312.html#!comments
http://www.cnblogs.com/apprenticeship/p/4215755.html
http://www.cnblogs.com/cmt/p/markdown-latex.html
http://blog.sina.com.cn/s/blog_5e16f1770100gror.html