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[求助] 求证:\(\left(2\sin\frac{\pi}{4}\right)^2=2\)

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发表于 2017-11-5 12:04:06 | 显示全部楼层 |阅读模式

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求证:
\(\left(2\sin\frac{\pi}{4}\right)^2=2\)
\(\left(2\sin\frac{\pi}{6}×2\sin\frac{2\pi}{6}\right)^2=3\)
\(\left(2\sin\frac{\pi}{8}×2\sin\frac{2\pi}{8}×2\sin\frac{3\pi}{8}\right)^2=4\)
\(\left(2\sin\frac{\pi}{10}×2\sin\frac{2\pi}{10}×2\sin\frac{3\pi}{10}×2\sin\frac{4\pi}{10}\right)^2=5\)
\(\left(2\sin\frac{\pi}{12}×2\sin\frac{2\pi}{12}×2\sin\frac{3\pi}{12}×2\sin\frac{4\pi}{12}×2\sin\frac{5\pi}{12}\right)^2=6\)
\(\left(2\sin\frac{\pi}{14}×2\sin\frac{2\pi}{14}×2\sin\frac{3\pi}{14}×2\sin\frac{4\pi}{14}×2\sin\frac{5\pi}{14}×2\sin\frac{6\pi}{14}\right)^2=7\)
\(\left(2\sin\frac{\pi}{16}×2\sin\frac{2\pi}{16}×2\sin\frac{3\pi}{16}×2\sin\frac{4\pi}{16}×2\sin\frac{5\pi}{16}×2\sin\frac{6\pi}{16}×2\sin\frac{7\pi}{16}\right)^2=8\)
\(\left(2\sin\frac{\pi}{18}×2\sin\frac{2\pi}{18}×2\sin\frac{3\pi}{18}×2\sin\frac{4\pi}{18}×2\sin\frac{5\pi}{18}×2\sin\frac{6\pi}{18}×2\sin\frac{7\pi}{18}×2\sin\frac{8\pi}{18}\right)^2=9\)
.............
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2017-11-8 19:17:45 | 显示全部楼层
求证:
   \(\D\left(2\sin\frac{1\pi}{4}\right)^2=2\)
   \(\D\left(2\sin\frac{2\pi}{6}\right)^2=3\)
   \(\D\left(2\sin\frac{2\pi}{8}×2\sin\frac{2\pi}{8}\right)^2=4\)
   \(\D\left(2\sin\frac{2\pi}{10}×2\sin\frac{4\pi}{10}\right)^2=5\)
   \(\D\left(2\sin\frac{2\pi}{12}×2\sin\frac{4\pi}{12}×2\sin\frac{3\pi}{12}\right)^2=6\)
   \(\D\left(2\sin\frac{2\pi}{14}×2\sin\frac{4\pi}{14}×2\sin\frac{6\pi}{14}\right)^2=7\)
   \(\D\left(2\sin\frac{2\pi}{16}×2\sin\frac{4\pi}{16}×2\sin\frac{6\pi}{16}×2\sin\frac{4\pi}{16}\right)^2=8\)
   \(\D\left(2\sin\frac{2\pi}{18}×2\sin\frac{4\pi}{18}×2\sin\frac{6\pi}{18}×2\sin\frac{8\pi}{18}\right)^2=9\)
   \(\D\left(2\sin\frac{2\pi}{20}×2\sin\frac{4\pi}{20}×2\sin\frac{6\pi}{20}×2\sin\frac{8\pi}{20}×2\sin\frac{5\pi}{20}\right)^2=10\)
   \(\D\left(2\sin\frac{2\pi}{22}×2\sin\frac{4\pi}{22}×2\sin\frac{6\pi}{22}×2\sin\frac{8\pi}{22}×2\sin\frac{10\pi}{22}\right)^2=11\)
   \(\D\left(2\sin\frac{2\pi}{24}×2\sin\frac{4\pi}{24}×2\sin\frac{6\pi}{24}×2\sin\frac{8\pi}{24}×2\sin\frac{10\pi}{24}×2\sin\frac{6\pi}{24}\right)^2=12\)
   \(\D\left(2\sin\frac{2\pi}{26}×2\sin\frac{4\pi}{26}×2\sin\frac{6\pi}{26}×2\sin\frac{8\pi}{26}×2\sin\frac{10\pi}{26}×2\sin\frac{12\pi}{26}\right)^2=13\)
   \(\D\left(2\sin\frac{2\pi}{28}×2\sin\frac{4\pi}{28}×2\sin\frac{6\pi}{28}×2\sin\frac{8\pi}{28}×2\sin\frac{10\pi}{28}×2\sin\frac{12\pi}{28}×2\sin\frac{7\pi}{28}\right)^2=14\)
   \(\D\left(2\sin\frac{2\pi}{30}×2\sin\frac{4\pi}{30}×2\sin\frac{6\pi}{30}×2\sin\frac{8\pi}{30}×2\sin\frac{10\pi}{30}×2\sin\frac{12\pi}{30}×2\sin\frac{14\pi}{30}\right)^2=15\)
   .............

毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2017-11-8 22:30:01 | 显示全部楼层
利用对数,正弦的对数按以下关系变形。
\(\log (\sin (\theta ))=-i \theta +\log \left(1-e^{i \theta }\right)+\log \left(1+e^{i \theta }\right)+\frac{i \pi }{2}-\log (2)\)

点评

谢谢zeroieme!谢谢大家!我慢慢的还是琢磨。  发表于 2017-11-10 14:52
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2017-11-10 14:44:07 | 显示全部楼层
本帖最后由 王守恩 于 2017-11-10 16:24 编辑

    \(\D \frac{1}{\sin\frac{\pi}{4}\ \sin\frac{2\pi}{4}}+\frac{1}{\sin\frac{2\pi}{4}\ \sin\frac{3\pi}{4}}=\frac{2\cos\frac{\pi}{4}}{\left(\sin\frac{\pi}{4}\right)^2}\)

    \(\D \frac{1}{\sin\frac{\pi}{6}\ \sin\frac{2\pi}{6}}+\frac{1}{\sin\frac{2\pi}{6}\ \sin\frac{3\pi}{6}}+\frac{1}{\sin\frac{3\pi}{6}\ \sin\frac{4\pi}{6}}+\frac{1}{\sin\frac{4\pi}{6}\ \sin\frac{5\pi}

{6}}=\frac{2\cos\frac{\pi}{6}}{\left(\sin\frac{\pi}{6}\right)^2}\)

    \(\D \frac{1}{\sin\frac{\pi}{8}\ \sin\frac{2\pi}{8}}+\frac{1}{\sin\frac{2\pi}{8}\ \sin\frac{3\pi}{8}}+\frac{1}{\sin\frac{3\pi}{8}\ \sin\frac{4\pi}{8}}+\frac{1}{\sin\frac{4\pi}{8}\ \sin\frac{5\pi}{8}}+\frac{1}

{\sin\frac{5\pi}{8}\ \sin\frac{6\pi}{8}}+\frac{1}{\sin\frac{6\pi}{8}\ \sin\frac{7\pi}{8}}=\frac{2\cos\frac{\pi}{8}}{\left(\sin\frac{\pi}{8}\right)^2}\)

    \(\D \frac{1}{\sin\frac{\pi}{10}\ \sin\frac{2\pi}{10}}+\frac{1}{\sin\frac{2\pi}{10}\ \sin\frac{3\pi}{10}}+\frac{1}{\sin\frac{3\pi}{10}\ \sin\frac{4\pi}{10}}+\cdots+\frac{1}{\sin\frac{6\pi}{10}\

\sin\frac{7\pi}

{10}}+\frac{1}{\sin\frac{7\pi}{10}\ \sin\frac{8\pi}{10}}+\frac{1}{\sin\frac{8\pi}{10}\ \sin\frac{9\pi}{10}}=\frac{2\cos\frac{\pi}{10}}{\left(\sin\frac{\pi}{10}\right)^2}\)

    \(\D \frac{1}{\sin\frac{\pi}{12}\ \sin\frac{2\pi}{12}}+\frac{1}{\sin\frac{2\pi}{12}\ \sin\frac{3\pi}{12}}+\frac{1}{\sin\frac{3\pi}{12}\ \sin\frac{4\pi}{12}}+\cdots+\frac{1}{\sin\frac{8\pi}{12}\

\sin\frac{9\pi}

{12}}+\frac{1}{\sin\frac{9\pi}{12}\ \sin\frac{10\pi}{12}}+\frac{1}{\sin\frac{10\pi}{12}\ \sin\frac{11\pi}{12}}=\frac{2\cos\frac{\pi}{12}}{\left(\sin\frac{\pi}{12}\right)^2}\)

   \(\D \frac{1}{\sin\frac{\pi}{14}\ \sin\frac{2\pi}{14}}+\frac{1}{\sin\frac{2\pi}{14}\ \sin\frac{3\pi}{14}}+\frac{1}{\sin\frac{3\pi}{14}\ \sin\frac{4\pi}{14}}+\cdots+\frac{1}{\sin\frac{10\pi}{14}\

\sin\frac{11\pi}

{14}}+\frac{1}{\sin\frac{11\pi}{14}\ \sin\frac{12\pi}{14}}+\frac{1}{\sin\frac{12\pi}{14}\ \sin\frac{13\pi}{14}}=\frac{2\cos\frac{\pi}{14}}{\left(\sin\frac{\pi}{14}\right)^2}\)

毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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