讨论:三角恒等式

本条目属于下列维基专题范畴：

（获评未评级、中重要度）

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中	根据专题重要度评级标准，本条目已评为中重要度。

基本关系

调整前

函数	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 \$

调整后

函数	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 \$

明明照关系去排表格是一张彼此对衬的表格，是谁这么想不开，把cot与csc的位置对调？-P1ayer (留言) 2010年4月1日 (四) 21:56 (UTC)[回复]

各位维基人：

我刚刚修改了三角恒等式中的2个外部链接，请大家仔细检查我的编辑。如果您有疑问，或者需要让机器人忽略某个链接甚至整个页面，请访问这个简单的FAQ获取更多信息。我进行了以下修改：

有关机器人修正错误的详情请参阅FAQ。