公式 - OLENCER's blog.

求和

$ \sum \limits_{i = 1} ^{n} {i} = \frac{n(n + 1)}{2} $

$$
\begin{align}
&\sum \limits_{i = 1} ^{n} {i} = 1 + 2 + 3 + ··· + n \
&\sum \limits_{i = n} ^{1} {i} = n + ··· + 3 + 2 + 1 \
2 &\sum \limits_{i = 1} ^{n} {i} = n(n - 1) \
&\sum \limits_{i = 1} ^{n} {i} = \frac{n(n + 1)}{2}
\end{align}
$$

$ \sum \limits_{i = 1} ^{n} {i ^{2} } = \frac{n(n + 1)(2n + 1)}{6} $

$$
\begin{align}
&(n + 1)^{3} - n^{3} = 3n^2 + 3n + 1 \
&n^{3} - (n - 1)^{3} = 3(n - 1)^2 + 3(n - 1) + 1 \ \
& ······ \ \
&3^{3} - 2^{3} = 3 \times 2^2 + 3 \times 2 + 1 \
&2^{2} - 1^{3} = 3 \times 1^2 + 3 \times 1 + 1 \ \
&(n + 1)^{3} - 1^{3} = 3 \sum \limits_{i = 1} ^{n} {i ^{2} } + 3\sum \limits_{i = 1} ^{n} {i} + n \
&(n^{3} + 3n^{2} + 3n + 1) - 1 = 3 \sum \limits_{i = 1} ^{n} {i ^{2} } + \frac{3n(n + 1)}{2} + n \
&3 \sum \limits_{i = 1} ^{n} {i ^{2} } = n^{3} + 3n^{2} + 3n - [\frac{3n(n + 1)}{2} + n] \
&\sum \limits_{i = 1} ^{n} {i ^{2} } = \frac{n^{3} }{3} + n^{2} + n - \frac{n(n + 1)}{2} - \frac{n}{3} \
&\sum \limits_{i = 1} ^{n} {i ^{2} } = \frac{n^{3} }{3} + n^{2} + \frac{2n}{3} - \frac{n^{2} }{2} - \frac{n}{2} \
&\sum \limits_{i = 1} ^{n} {i ^{2} } = \frac{n^{3} }{3} + \frac{n^{2} }{2} + \frac{n}{6} = \frac{2n^3 + 3n^{2} + n}{6} = \frac{n(2n^2 + 3n + 1)}{6} = \frac{n(n + 1)(2n + 1)}{6} \
\end{align}
$$

$ \sum \limits_{i = 1} ^{n} {i ^{3} } = [\frac{n(n + 1)}{2}]^{2} $

$$
\begin{align}
&(n + 1)^{4} - n^{4} = 4n^3 + 6n^2 + 4n + 1 \
&n^{4} - (n - 1)^{4} = 4(n - 1)^{3} + 6(n - 1)^{2} + 4(n - 1) + 1 \ \
& ······ \ \
&3^{4} - 2^{4} = 4 \times 2^3 + 6 \times 2^{2} + 4 \times 2 + 1 \
&2^{4} - 1^{4} = 4 \times 1^3 + 6 \times 1^{2} + 4 \times 1 + 1 \ \
&(n + 1)^{4} - 1^{4} = 4 \sum \limits_{i = 1} ^{n} {i ^{3} } + 6 \sum \limits_{i = 1} ^{n} {i^{2} } + 4\sum \limits_{i = 1} ^{n} {i} + n \
&(n^{4} + 4n^3 + 6n^2 + 4n + 1 ) - 1 = 4\sum \limits_{i = 1} ^{n} {i ^{3} } + 6[\frac{n(n + 1)(2n + 1)}{6}] + 4[\frac{n(n + 1)}{2}] - n \
&4\sum \limits_{i = 1} ^{n} {i ^{3} } = n^{4} + 4n^3 + 6n^2 + 4n - n(n + 1)(2n + 1) - 2n(n + 1) - n \
&\sum \limits_{i = 1} ^{n} {i ^{3} } = \frac{n^{4} }{4} + n^3 + \frac{3n^2}{2} + n - \frac{n(n + 1)(2n + 1)}{4} - \frac{n(n + 1)}{2} - \frac{n}{4} \
&\sum \limits_{i = 1} ^{n} {i ^{3} } = \frac{n^{4} }{4} + n^3 + \frac{3n^2}{2} + n - \frac{n ^ {3} }{2} - \frac{n^{2} }{4} - \frac{n ^ {2} }{2} - \frac{n}{4} - \frac{n ^ {2} }{2} - \frac{n}{2} - \frac{n}{4} \
&\sum \limits_{i = 1} ^{n} {i ^{3} } = \frac{n^{4} }{4} + \frac{n^3}{2} + \frac{n^2}{4} = \frac{n ^4 + 2n^3 + n^2}{4} = \frac{n ^ 2(n ^2 + 2n + 1)}{4} = \frac{n ^ 2(n + 1)^2}{4} = [\frac{n(n + 1)}{2}]^{2} \
\end{align}
$$

和差化积

$ \sin{\alpha} + \sin{\beta} = 2 \sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} })$

$$
\begin{align}
&\sin{\alpha} + \sin{\beta} \
=\quad &\sin{(\frac{\alpha + \beta}{2} + \frac{\alpha - \beta}{2} }) + \sin{(\frac{\alpha + \beta}{2} - \frac{\alpha - \beta}{2} }) \
=\quad &[\sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) + \cos{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] + [\sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) - \cos{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] \
=\quad &[\sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) + \sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{-\alpha + \beta}{2} })] + [\cos{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} }) - \cos{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] \
=\quad &2\sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} })
\end{align}
$$

$ \sin{\alpha} - \sin{\beta} = 2 \cos{(\frac{\alpha + \beta}{2} }) \sin{(\frac{\alpha - \beta}{2})}$

$$
\begin{align}
&\sin{\alpha} - \sin{\beta} \
=\quad &\sin{(\frac{\alpha + \beta}{2} + \frac{\alpha - \beta}{2} }) - \sin{(\frac{\alpha + \beta}{2} - \frac{\alpha - \beta}{2} }) \
=\quad &[\sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) + \cos{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] - [\sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) - \cos{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] \
=\quad &[\sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) - \sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{-\alpha + \beta}{2} })] + [\cos{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} }) + \cos{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] \
=\quad &2 \cos{(\frac{\alpha + \beta}{2} }) \sin{(\frac{\alpha - \beta}{2})}
\end{align}
$$

$ \cos{\alpha} + \cos{\beta} = 2 \cos{(\frac{\alpha + \beta}{2} }) \cos{(\frac{\alpha - \beta}{2})}$

$$
\begin{align}
&\cos{\alpha} + \cos{\beta} \
=\quad &\cos{(\frac{\alpha + \beta}{2} + \frac{\alpha - \beta}{2} }) + \cos{(\frac{\alpha + \beta}{2} - \frac{\alpha - \beta}{2} }) \
=\quad &[\cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) - \sin{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] + [\cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) + \sin{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] \
=\quad &[\cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) + \cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{-\alpha + \beta}{2} })] + [\sin{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} }) - \sin{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] \
=\quad &2 \cos{(\frac{\alpha + \beta}{2} }) \cos{(\frac{\alpha - \beta}{2})}
\end{align}
$$

$ \cos{\alpha} - \cos{\beta} = -2 \sin{(\frac{\alpha + \beta}{2} }) \sin{(\frac{\alpha - \beta}{2})} $

$$
\begin{align}
&\cos{\alpha} - \cos{\beta} \
=\quad &\cos{(\frac{\alpha + \beta}{2} + \frac{\alpha - \beta}{2} }) - \cos{(\frac{\alpha + \beta}{2} - \frac{\alpha - \beta}{2} }) \
=\quad &[\cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) - \sin{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] - [\cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) + \sin{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] \
=\quad &[\cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} }) - \cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{-\alpha + \beta}{2} })] - [\sin{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} }) + \sin{(\frac{\alpha + \beta}{2})} \sin{(\frac{\alpha - \beta}{2} })] \
=\quad &-2 \sin{(\frac{\alpha + \beta}{2} }) \sin{(\frac{\alpha - \beta}{2})}
\end{align}
$$

$ \tan{\alpha} \pm \tan{\beta} = \frac{\sin{(\alpha \pm \beta)} }{\cos{\alpha} \cos{\beta} }$

$$
\begin{align}
&\tan{\alpha} \pm \tan{\beta} \
=\quad &\frac{\sin{\alpha} }{\cos{\alpha} } \pm \frac{\sin{\beta} }{\cos{\beta} } \
=\quad &\frac{\sin{\alpha} \cos{\beta} \pm \cos{\alpha} \sin{\beta} }{\cos{\alpha} \cos{\beta} } \
=\quad &\frac{\sin{(\alpha \pm \beta)} }{\cos{\alpha} \cos{\beta} }
\end{align}
$$

平方形式的和差化积公式

$ \sin^{2}{\alpha} - \sin^{2}{\beta} = \sin{(\alpha +\beta)} \sin{(\alpha -\beta)} $

$$
\begin{align}
&\sin^{2}{\alpha} - \sin^{2}{\beta} \
=\quad &(\sin{\alpha} + \sin{\beta}) (\sin{\alpha} - \sin{\beta}) \
=\quad &[2 \sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} })] [2 \cos{(\frac{\alpha + \beta}{2} }) \sin{(\frac{\alpha - \beta}{2})]} \
=\quad &[2 \sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha + \beta}{2} })] [2 \sin{(\frac{\alpha - \beta}{2} }) \cos{(\frac{\alpha - \beta}{2})]} \
=\quad &\sin{(\alpha +\beta)} \sin{(\alpha -\beta)}
\end{align}
$$

$ \cos^{2}{\alpha} - \cos^{2}{\beta} = - \sin{(\alpha +\beta)} \sin{(\alpha -\beta)} $

$$
\begin{align}
&\cos^{2}{\alpha} - \cos^{2}{\beta} \
=\quad &(\cos{\alpha} + \cos{\beta}) (\cos{\alpha} - \cos{\beta}) \
=\quad &[2 \cos{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha - \beta}{2} })] [-2 \sin{(\frac{\alpha + \beta}{2} }) \sin{(\frac{\alpha - \beta}{2})]} \
=\quad &- [2 \sin{(\frac{\alpha + \beta}{2})} \cos{(\frac{\alpha + \beta}{2} })] [2 \sin{(\frac{\alpha - \beta}{2} }) \cos{(\frac{\alpha - \beta}{2})]} \
=\quad &- \sin{(\alpha +\beta)} \sin{(\alpha -\beta)}
\end{align}
$$

$ \cos^{2}{\alpha} - \sin^{2}{\beta} = \cos{(\alpha +\beta)} \cos{(\alpha -\beta)} $

$$
\begin{align}
&\cos^{2}{\alpha} - \sin^{2}{\beta} \
=\quad &(\cos^2{\alpha} - \cos^2{\alpha} \sin^2{\beta}) - (\sin^{2}{\beta} - \cos^2{\alpha} \sin^2{\beta}) \
=\quad &[\cos^2{\alpha} (1 - \sin^2{\beta})] - [\sin^{2}{\beta}(1 - \cos^2{\alpha})] \
=\quad &(\cos^2{\alpha} \cos^2{\beta}) - (\sin^{2}{\alpha} \cos^2{\beta} ) \
=\quad &(\cos{\alpha} \cos{\beta})^2 + \sin{\alpha} \sin{\beta} \cos{\alpha} \cos{\beta} - \sin{\alpha} \sin{\beta} \cos{\alpha} \cos{\beta} - (\sin{\alpha} \cos{\beta} )^{2} \
=\quad &(cos{\alpha} cos{\beta} - \sin{\alpha} \sin{\beta}) (cos{\alpha} cos{\beta} + \sin{\alpha} \sin{\beta}) \
=\quad &\cos{(\alpha +\beta)} \cos{(\alpha -\beta)}
\end{align}
$$

$ \sin^{2}{\alpha} - \cos^{2}{\beta} = - \cos{(\alpha +\beta)} \cos{(\alpha -\beta)} $

$$
\begin{align}
&\cos^{2}{\alpha’} - \sin^{2}{\beta’} = \cos{(\alpha’ +\beta’)} \cos{(\alpha’ -\beta’)} \
&\sin^{2}{\beta’} - \cos^{2}{\alpha’} = - \cos{(\alpha’ +\beta’)} \cos{(\alpha’ -\beta’)} \
&\sin^{2}{\beta’} - \cos^{2}{\alpha’} = - \cos{(\beta’ +\alpha’)} \cos{(\beta’ -\alpha’)} \ \

&令 \alpha = \beta', \beta = \alpha', 则 \\
&\sin^{2}{\alpha} - \cos^{2}{\beta} = - \cos{(\alpha +\beta)} \cos{(\alpha -\beta)}

\end{align}
$$