討論:三角恆等式

本條目屬於下列維基專題範疇：

（獲評未評級、中重要度）

	數學主題閱論編本條目屬於數學專題範疇，該專題旨在改善中文維基百科數學類內容。如果您有意參與，請瀏覽專題主頁、參與討論，並完成相應的開放性任務。
未評	根據專題品質評級標準，本條目尚未接受評級。
中	根據專題重要度評級標準，本條目已評為中重要度。

基本關係

調整前

函數	sin	cos	tan	csc	sec	cot
$\sin \theta =$	$\sin \theta \$	${\sqrt {1-\cos ^{2}\theta }}$	${\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}$	${\frac {1}{\csc \theta }}$	${\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}$	${\frac {1}{\sqrt {1+\cot ^{2}\theta }}}$
$\cos \theta =$	${\sqrt {1-\sin ^{2}\theta }}$	$\cos \theta \$	${\frac {1}{\sqrt {1+\tan ^{2}\theta }}}$	${\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}$	${\frac {1}{\sec \theta }}$	${\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}$
$\tan \theta =$	${\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}$	${\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}$	$\tan \theta \$	${\frac {1}{\sqrt {\csc ^{2}\theta -1}}}$	${\sqrt {\sec ^{2}\theta -1}}$	${\frac {1}{\cot \theta }}$
$\csc \theta =$	${1 \over \sin \theta }$	${1 \over {\sqrt {1-\cos ^{2}\theta }}}$	${{\sqrt {1+\tan ^{2}\theta }} \over \tan \theta }$	$\csc \theta \$	${\sec \theta \over {\sqrt {\sec ^{2}\theta -1}}}$	${\sqrt {1+\cot ^{2}\theta }}$
$\sec \theta =$	${1 \over {\sqrt {1-\sin ^{2}\theta }}}$	${1 \over \cos \theta }$	${\sqrt {1+\tan ^{2}\theta }}$	${\csc \theta \over {\sqrt {\csc ^{2}\theta -1}}}$	$\sec \theta \$	${{\sqrt {1+\cot ^{2}\theta }} \over \cot \theta }$
$\cot \theta =$	${{\sqrt {1-\sin ^{2}\theta }} \over \sin \theta }$	${\cos \theta \over {\sqrt {1-\cos ^{2}\theta }}}$	${1 \over \tan \theta }$	${\sqrt {\csc ^{2}\theta -1}}$	${1 \over {\sqrt {\sec ^{2}\theta -1}}}$	$\cot \theta \$

調整後

函數	sin	cos	tan	cot	sec	csc
$\sin \theta =$	$\sin \theta \$	${\sqrt {1-\cos ^{2}\theta }}$	${\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}$	${\frac {1}{\sqrt {1+\cot ^{2}\theta }}}$	${\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}$	${\frac {1}{\csc \theta }}$
$\cos \theta =$	${\sqrt {1-\sin ^{2}\theta }}$	$\cos \theta \$	${\frac {1}{\sqrt {1+\tan ^{2}\theta }}}$	${\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}$	${\frac {1}{\sec \theta }}$	${\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}$
$\tan \theta =$	${\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}$	${\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}$	$\tan \theta \$	${\frac {1}{\cot \theta }}$	${\sqrt {\sec ^{2}\theta -1}}$	${\frac {1}{\sqrt {\csc ^{2}\theta -1}}}$
$\cot \theta =$	${{\sqrt {1-\sin ^{2}\theta }} \over \sin \theta }$	${\cos \theta \over {\sqrt {1-\cos ^{2}\theta }}}$	${1 \over \tan \theta }$	$\cot \theta \$	${1 \over {\sqrt {\sec ^{2}\theta -1}}}$	${\sqrt {\csc ^{2}\theta -1}}$
$\sec \theta =$	${1 \over {\sqrt {1-\sin ^{2}\theta }}}$	${1 \over \cos \theta }$	${\sqrt {1+\tan ^{2}\theta }}$	${{\sqrt {1+\cot ^{2}\theta }} \over \cot \theta }$	$\sec \theta \$	${\csc \theta \over {\sqrt {\csc ^{2}\theta -1}}}$
$\csc \theta =$	${1 \over \sin \theta }$	${1 \over {\sqrt {1-\cos ^{2}\theta }}}$	${{\sqrt {1+\tan ^{2}\theta }} \over \tan \theta }$	${\sqrt {1+\cot ^{2}\theta }}$	${\sec \theta \over {\sqrt {\sec ^{2}\theta -1}}}$	$\csc \theta \$

明明照關係去排表格是一張彼此對襯的表格，是誰這麼想不開，把cot與csc的位置對調？-P1ayer (留言) 2010年4月1日 (四) 21:56 (UTC)[回覆]

各位維基人：

我剛剛修改了三角恆等式中的2個外部連結，請大家仔細檢查我的編輯。如果您有疑問，或者需要讓機械人忽略某個連結甚至整個頁面，請訪問這個簡單的FAQ獲取更多信息。我進行了以下修改：

有關機械人修正錯誤的詳情請參閱FAQ。