\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}

\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}
 

參考鏈接如下：
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

如何在博客園中插入公式