是不是要用单位根？

hbhuang1983

$$\sin \frac{2\pi}7+\sin \frac{3\pi}7-\sin \frac\pi7$$

色k

玩积化和差而已
\begin{align*}
&\left( \sin \frac{2\pi}7+\sin \frac{3\pi}7-\sin \frac\pi7 \right)^2 \\
={}&\sin^2\frac\pi7+\sin^2\frac{2\pi}7+\sin^2\frac{3\pi}7+2\sin \frac{2\pi}7\sin \frac{3\pi}7-2\sin \frac\pi7\sin \frac{2\pi}7-2\sin \frac\pi7\sin \frac{3\pi}7 \\
={}&\frac32-\frac12\left( \cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7 \right)-\cos \frac{5\pi}7+\cos \frac\pi7+\cos \frac{3\pi }7-\cos \frac\pi7+\cos \frac{4\pi}7-\cos \frac{2\pi}7 \\
={}&\frac32-\frac12\left( \cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7 \right),
\end{align*}
因为
\begin{align*}
\left( \cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7 \right)\sin \frac\pi7
&=\frac12\left( \sin \frac{3\pi}7-\sin \frac\pi7+\sin \frac{5\pi}7-\sin \frac{3\pi}7+\sin \pi-\sin \frac{5\pi}7 \right)\\
&=-\frac12\sin \frac\pi7,
\end{align*}
所以
\[\cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7=-\frac12,\]
代回去开方即得
\[\sin \frac{2\pi}7+\sin \frac{3\pi}7-\sin \frac\pi7=\frac{\sqrt7}2.\]

色k

再来一个坑爹解法：
记
\[f=\sin \frac{2\pi}7+\sin \frac{3\pi}7-\sin \frac\pi7,\]
则
\[2f=\sin \frac\pi7+\sin \frac{2\pi}7+\cdots +\sin \frac{6\pi}7-4\sin \frac\pi7,\]
类似上面那样积化和差，有
\[\left( \sin \frac\pi7+\sin \frac{2\pi}7+\cdots +\sin \frac{6\pi}7 \right)\sin \frac\pi{14}=\frac12\left( \cos \frac\pi{14}-\cos \frac{13\pi}{14} \right)=\cos \frac\pi{14},\]
则
\[\sin \frac\pi7+\sin \frac{2\pi}7+\cdots +\sin \frac{6\pi}7=\cot \frac\pi{14},\]
所以
\[2f=\cot \frac\pi{14}-4\sin \frac\pi7.\]

为方便书写，记 $t=\cot(\pi/14)$，因为 $\cot(7\pi/14)=0$，则由 $n$ 倍角余切公式可得
\[C_7^0t^7-C_7^2t^5+C_7^4t^3-C_7^6t=0,\]
化简即得
\[t^6=7(3t^4-5t^2+1),\]
由万能公式有
\[2f=t-\frac{8t}{1+t^2}=\frac{t^3-7t}{1+t^2},\]
平方得
\[4f^2=\frac{t^6-14t^4+49t^2}{t^4+2t^2+1}=\frac{7(3t^4-5t^2+1)-14t^4+49t^2}{t^4+2t^2+1}=7,\]
即
\[f=\frac{\sqrt7}2.\]

isee

玩积化和差而已
\begin{align*}
&\left( \sin \frac{2\pi}7+\sin \frac{3\pi}7-\sin \frac\pi7 \right)^2 \ ...
色k 发表于 2016-4-1 00:48

源自知乎提问

原来在你的过程中就把这题解了


题：证： $\cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{8\pi}7=-\frac 12. $

一个字：失传的积化和差

\begin{gather*} \left( \cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{8\pi}7 \right)\sin \frac\pi7\\[1em] =\left( \cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{6\pi}7 \right)\sin \frac\pi7\\[1em] =\sin \frac\pi7 \cos \frac{2\pi}7+\sin \frac\pi7\cos \frac{4\pi}7+\sin \frac\pi7\cos \frac{6\pi}7\\[1em] =\frac12\left( \sin \frac{3\pi}7-\sin \frac\pi7+\sin \frac{5\pi}7-\sin \frac{3\pi}7+\sin \pi-\sin \frac{5\pi}7 \right)\\[1em] =-\frac12\sin \frac\pi7\\[1em] \therefore {~}\cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{8\pi}7=-\frac 12.  \end{gather*}

PS：链接中有复数法，几何法等也是精彩的

isee

源自

原来在你的过程中就把这题解了


题：证： $\cos \frac{2\pi}7+\cos \frac{4\pi}7+\cos \frac{8\pi ...
isee 发表于 2022-2-24 00:25

也就是单位根（复数法）