在數學中，三角恆等式是對出現的所有值都為實變量，涉及到三角函數的等式。這些恆等式在表達式中有些三角函數需要簡化的時候是很有用的。一個重要應用是非三角函數的積分：一個常用技巧是首先使用使用三角函數的代換規則，則通過三角恆等式可簡化結果的積分。 在幾何上依據以O為中心的單位圓可以構造角θ的很多三角函數 三角函數示意圖 由於已知的技術原因，圖表暫時不可用。帶來不便，我們深表歉意。 幾個三角函數的圖形，分別為正弦、餘弦、正切、餘切、正割、餘割和正矢。配色與上圖相同 單位圓的角度 符號 為了避免由於 sin − 1 x {\displaystyle \sin ^{-1}x} 的不同意思所帶來的混淆，我們經常用下列兩個表格來表示三角函數的倒數和反函數。另外在表示餘割函數時，' csc {\displaystyle \csc } '有時會寫成比較長的' c o s e c {\displaystyle \mathrm {cosec} } '。 更多資訊 函數, 反函數 ... 函數 反函數 倒數 中文 全寫 簡寫 中文 全寫 簡寫 中文 全寫 簡寫 正弦 sine sin 反正弦 arcsine arcsin 餘割 cosecant csc 餘弦 cosine cos 反餘弦 arccosine arccos 正割 secant sec 正切 tangent tan 反正切 arctangent arctan 餘切 cotangent cot 餘切 cotangent cot 反餘切 arccotangent arccot 正切 tangent tan 正割 secant sec 反正割 arcsecant arcsec 餘弦 cosine cos 餘割 cosecant csc 反餘割 arccosecant arccsc 正弦 sine sin 關閉 不同的角度度量適合於不同的情況。本表展示最常用的系統。弧度是預設的角度量並用在指數函數中。所有角度度量都是無單位的。另外在計算機中角度的符號為D，弧度的符號為R，梯度的符號為G。 更多資訊 , ... 相同角度的轉換表 角度單位 值 計算機中代號 轉 0 {\displaystyle 0} 1 12 {\displaystyle {\frac {1}{12}}} 1 8 {\displaystyle {\frac {1}{8}}} 1 6 {\displaystyle {\frac {1}{6}}} 1 4 {\displaystyle {\frac {1}{4}}} 1 2 {\displaystyle {\frac {1}{2}}} 3 4 {\displaystyle {\frac {3}{4}}} 1 {\displaystyle 1} 無 角度 0 ∘ {\displaystyle 0^{\circ }} 30 ∘ {\displaystyle 30^{\circ }} 45 ∘ {\displaystyle 45^{\circ }} 60 ∘ {\displaystyle 60^{\circ }} 90 ∘ {\displaystyle 90^{\circ }} 180 ∘ {\displaystyle 180^{\circ }} 270 ∘ {\displaystyle 270^{\circ }} 360 ∘ {\displaystyle 360^{\circ }} D 弧度 0 {\displaystyle 0} π 6 {\displaystyle {\frac {\pi }{6}}} π 4 {\displaystyle {\frac {\pi }{4}}} π 3 {\displaystyle {\frac {\pi }{3}}} π 2 {\displaystyle {\frac {\pi }{2}}} π {\displaystyle \pi } 3 π 2 {\displaystyle {\frac {3\pi }{2}}} 2 π {\displaystyle 2\pi } R 梯度 0 g {\displaystyle 0^{g}} 33 1 3 g {\displaystyle 33{\frac {1}{3}}^{g}} 50 g {\displaystyle 50^{g}} 66 2 3 g {\displaystyle 66{\frac {2}{3}}^{g}} 100 g {\displaystyle 100^{g}} 200 g {\displaystyle 200^{g}} 300 g {\displaystyle 300^{g}} 400 g {\displaystyle 400^{g}} G 關閉 基本關係 三角函數間的關係，可分成正函數和餘函數 畢達哥拉斯三角恆等式如下： sin 2 θ + cos 2 θ = 1 {\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1\,} tan 2 θ + 1 = sec 2 θ {\displaystyle \tan ^{2}\theta +1\,=\sec ^{2}\theta } 1 + cot 2 θ = csc 2 θ {\displaystyle 1\,+\cot ^{2}\theta =\csc ^{2}\theta } 由上面的平方關係加上三角函數的基本定義，可以導出下面的表格，即每個三角函數都可以用其他五個表達。（嚴謹地說，所有根號前都應根據實際情況添加正負號） 更多資訊 , ... 函數 sin {\displaystyle \sin } cos {\displaystyle \cos } tan {\displaystyle \tan } cot {\displaystyle \cot } sec {\displaystyle \sec } csc {\displaystyle \csc } sin θ {\displaystyle \sin \theta } sin θ {\displaystyle \sin \theta \ } 1 − cos 2 θ {\displaystyle {\sqrt {1-\cos ^{2}\theta }}} tan θ 1 + tan 2 θ {\displaystyle {\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}} 1 1 + cot 2 θ {\displaystyle {\frac {1}{\sqrt {1+\cot ^{2}\theta }}}} sec 2 θ − 1 sec θ {\displaystyle {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}} 1 csc θ {\displaystyle {\frac {1}{\csc \theta }}} cos θ {\displaystyle \cos \theta } 1 − sin 2 θ {\displaystyle {\sqrt {1-\sin ^{2}\theta }}} cos θ {\displaystyle \cos \theta \ } 1 1 + tan 2 θ {\displaystyle {\frac {1}{\sqrt {1+\tan ^{2}\theta }}}} cot θ 1 + cot 2 θ {\displaystyle {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}} 1 sec θ {\displaystyle {\frac {1}{\sec \theta }}} csc 2 θ − 1 csc θ {\displaystyle {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}} tan θ {\displaystyle \tan \theta } sin θ 1 − sin 2 θ {\displaystyle {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}} 1 − cos 2 θ cos θ {\displaystyle {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}} tan θ {\displaystyle \tan \theta \ } 1 cot θ {\displaystyle {\frac {1}{\cot \theta }}} sec 2 θ − 1 {\displaystyle {\sqrt {\sec ^{2}\theta -1}}} 1 csc 2 θ − 1 {\displaystyle {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}} cot θ {\displaystyle \cot \theta } 1 − sin 2 θ sin θ {\displaystyle {{\sqrt {1-\sin ^{2}\theta }} \over \sin \theta }} cos θ 1 − cos 2 θ {\displaystyle {\cos \theta \over {\sqrt {1-\cos ^{2}\theta }}}} 1 tan θ {\displaystyle {1 \over \tan \theta }} cot θ {\displaystyle \cot \theta \ } 1 sec 2 θ − 1 {\displaystyle {1 \over {\sqrt {\sec ^{2}\theta -1}}}} csc 2 θ − 1 {\displaystyle {\sqrt {\csc ^{2}\theta -1}}} sec θ {\displaystyle \sec \theta } 1 1 − sin 2 θ {\displaystyle {1 \over {\sqrt {1-\sin ^{2}\theta }}}} 1 cos θ {\displaystyle {1 \over \cos \theta }} 1 + tan 2 θ {\displaystyle {\sqrt {1+\tan ^{2}\theta }}} 1 + cot 2 θ cot θ {\displaystyle {{\sqrt {1+\cot ^{2}\theta }} \over \cot \theta }} sec θ {\displaystyle \sec \theta \ } csc θ csc 2 θ − 1 {\displaystyle {\csc \theta \over {\sqrt {\csc ^{2}\theta -1}}}} csc θ {\displaystyle \csc \theta } 1 sin θ {\displaystyle {1 \over \sin \theta }} 1 1 − cos 2 θ {\displaystyle {1 \over {\sqrt {1-\cos ^{2}\theta }}}} 1 + tan 2 θ tan θ {\displaystyle {{\sqrt {1+\tan ^{2}\theta }} \over \tan \theta }} 1 + cot 2 θ {\displaystyle {\sqrt {1+\cot ^{2}\theta }}} sec θ sec 2 θ − 1 {\displaystyle {\sec \theta \over {\sqrt {\sec ^{2}\theta -1}}}} csc θ {\displaystyle \csc \theta \ } 關閉 其他函數的基本關係 正矢、餘矢、半正矢、半餘矢、外正割用於航行。例如半正矢可以計算球體上的兩個點之間的距離，但它們不常用。 更多資訊 , ... 名稱 函數 值[1] 正矢, versine versin θ {\displaystyle \operatorname {versin} \theta } vers θ {\displaystyle \operatorname {vers} \theta } ver θ {\displaystyle \operatorname {ver} \theta } 1 − cos θ {\displaystyle 1-\cos \theta } 餘的正矢, vercosine vercosin θ {\displaystyle \operatorname {vercosin} \theta } 1 + cos θ {\displaystyle 1+\cos \theta } 餘矢, coversine coversin θ {\displaystyle \operatorname {coversin} \theta } cvs θ {\displaystyle \operatorname {cvs} \theta } 1 − sin θ {\displaystyle 1-\sin \theta } 餘的餘矢, covercosine covercosin θ {\displaystyle \operatorname {covercosin} \theta } 1 + sin θ {\displaystyle 1+\sin \theta } 半正矢, haversine haversin θ {\displaystyle \operatorname {haversin} \theta } 1 − cos θ 2 {\displaystyle {\frac {1-\cos \theta }{2}}} 餘的半正矢, havercosine havercosin θ {\displaystyle \operatorname {havercosin} \theta } 1 + cos θ 2 {\displaystyle {\frac {1+\cos \theta }{2}}} 半餘矢, hacoversinecohaversine hacoversin θ {\displaystyle \operatorname {hacoversin} \theta } 1 − sin θ 2 {\displaystyle {\frac {1-\sin \theta }{2}}} 餘的半餘矢, hacovercosinecohavercosine hacovercosin θ {\displaystyle \operatorname {hacovercosin} \theta } 1 + sin θ 2 {\displaystyle {\frac {1+\sin \theta }{2}}} 外正割，exsecant exsec θ {\displaystyle \operatorname {exsec} \theta } sec θ − 1 {\displaystyle \sec \theta -1} 外餘割，excosecant excsc θ {\displaystyle \operatorname {excsc} \theta } csc θ − 1 {\displaystyle \csc \theta -1} 弦函數, chord crd θ {\displaystyle \operatorname {crd} \theta } 2 sin ( θ 2 ) {\displaystyle 2\sin \left({\frac {\theta }{2}}\right)} 純虛數指數函數, cosine and imaginary unit sine cis θ {\displaystyle \operatorname {cis} \theta } cos θ + i sin θ {\displaystyle \cos \theta +i\;\sin \theta } 輻角,Argument arg x {\displaystyle \arg x} Im ( ln x ) {\displaystyle \operatorname {Im} (\ln x)} 關閉 對稱、移位和周期 通過檢視單位圓，可確立三角函數的下列性質，這些性質也被稱為誘導公式： 維基教科書有誘導公式相關的教科書和手冊 對稱 當三角函數反射自某個特定的 θ {\displaystyle \theta } 值，結果經常是另一個其他三角函數。這導致了下列恆等式： 更多資訊 反射於 ... 反射於 θ = 0 {\displaystyle \theta =0} 反射於 θ = π 4 {\displaystyle \theta ={\tfrac {\pi }{4}}} 反射於 θ = π 2 {\displaystyle \theta ={\tfrac {\pi }{2}}} 反射於 θ = 3 π 4 {\displaystyle \theta ={\tfrac {3\pi }{4}}} sin ( 0 − θ ) = − sin θ cos ( 0 − θ ) = + cos θ tan ( 0 − θ ) = − tan θ cot ( 0 − θ ) = − cot θ sec ( 0 − θ ) = + sec θ csc ( 0 − θ ) = − csc θ {\displaystyle {\begin{aligned}\sin(0-\theta )&=-\sin \theta \\\cos(0-\theta )&=+\cos \theta \\\tan(0-\theta )&=-\tan \theta \\\cot(0-\theta )&=-\cot \theta \\\sec(0-\theta )&=+\sec \theta \\\csc(0-\theta )&=-\csc \theta \end{aligned}}} sin ( π 2 − θ ) = + cos θ cos ( π 2 − θ ) = + sin θ tan ( π 2 − θ ) = + cot θ cot ( π 2 − θ ) = + tan θ sec ( π 2 − θ ) = + csc θ csc ( π 2 − θ ) = + sec θ {\displaystyle {\begin{aligned}\sin({\tfrac {\pi }{2}}-\theta )&=+\cos \theta \\\cos({\tfrac {\pi }{2}}-\theta )&=+\sin \theta \\\tan({\tfrac {\pi }{2}}-\theta )&=+\cot \theta \\\cot({\tfrac {\pi }{2}}-\theta )&=+\tan \theta \\\sec({\tfrac {\pi }{2}}-\theta )&=+\csc \theta \\\csc({\tfrac {\pi }{2}}-\theta )&=+\sec \theta \end{aligned}}} sin ( π − θ ) = + sin θ cos ( π − θ ) = − cos θ tan ( π − θ ) = − tan θ cot ( π − θ ) = − cot θ sec ( π − θ ) = − sec θ csc ( π − θ ) = + csc θ {\displaystyle {\begin{aligned}\sin(\pi -\theta )&=+\sin \theta \\\cos(\pi -\theta )&=-\cos \theta \\\tan(\pi -\theta )&=-\tan \theta \\\cot(\pi -\theta )&=-\cot \theta \\\sec(\pi -\theta )&=-\sec \theta \\\csc(\pi -\theta )&=+\csc \theta \end{aligned}}} sin ( 3 π 2 − θ ) = − cos θ cos ( 3 π 2 − θ ) = − sin θ tan ( 3 π 2 − θ ) = + cot θ cot ( 3 π 2 − θ ) = + tan θ sec ( 3 π 2 − θ ) = − csc θ csc ( 3 π 2 − θ ) = − sec θ {\displaystyle {\begin{aligned}\sin({\tfrac {3\pi }{2}}-\theta )&=-\cos \theta \\\cos({\tfrac {3\pi }{2}}-\theta )&=-\sin \theta \\\tan({\tfrac {3\pi }{2}}-\theta )&=+\cot \theta \\\cot({\tfrac {3\pi }{2}}-\theta )&=+\tan \theta \\\sec({\tfrac {3\pi }{2}}-\theta )&=-\csc \theta \\\csc({\tfrac {3\pi }{2}}-\theta )&=-\sec \theta \end{aligned}}} 關閉 移位和周期 通過旋轉特定角度移位三角函數，經常可以找到更簡單的表達結果的不同的三角函數。例如通過旋轉 π 2 {\displaystyle {\tfrac {\pi }{2}}} 、 π {\displaystyle \pi } 和 2 π {\displaystyle 2\pi } 弧度移位函數。因爲這些函數的周期要麼是 π {\displaystyle \pi } 要麼是 2 π {\displaystyle 2\pi } ，所以新函數和沒有移位的舊函數完全一樣。 更多資訊 移位 ... 移位 π 2 {\displaystyle {\tfrac {\pi }{2}}} 移位 π {\displaystyle \pi } 移位 3 π 2 {\displaystyle {\tfrac {3\pi }{2}}} 移位 2 π {\displaystyle 2\pi } tan {\displaystyle \tan } 和 cot {\displaystyle \cot } 的周期 sin {\displaystyle \sin } , cos {\displaystyle \cos } , csc {\displaystyle \csc } 和 sec {\displaystyle \sec } 的周期 sin ( θ + π 2 ) = + cos θ cos ( θ + π 2 ) = − sin θ tan ( θ + π 2 ) = − cot θ cot ( θ + π 2 ) = − tan θ sec ( θ + π 2 ) = − csc θ csc ( θ + π 2 ) = + sec θ {\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {\pi }{2}})&=+\cos \theta \\\cos(\theta +{\tfrac {\pi }{2}})&=-\sin \theta \\\tan(\theta +{\tfrac {\pi }{2}})&=-\cot \theta \\\cot(\theta +{\tfrac {\pi }{2}})&=-\tan \theta \\\sec(\theta +{\tfrac {\pi }{2}})&=-\csc \theta \\\csc(\theta +{\tfrac {\pi }{2}})&=+\sec \theta \end{aligned}}} sin ( θ + π ) = − sin θ cos ( θ + π ) = − cos θ tan ( θ + π ) = + tan θ cot ( θ + π ) = + cot θ sec ( θ + π ) = − sec θ csc ( θ + π ) = − csc θ {\displaystyle {\begin{aligned}\sin(\theta +\pi )&=-\sin \theta \\\cos(\theta +\pi )&=-\cos \theta \\\tan(\theta +\pi )&=+\tan \theta \\\cot(\theta +\pi )&=+\cot \theta \\\sec(\theta +\pi )&=-\sec \theta \\\csc(\theta +\pi )&=-\csc \theta \end{aligned}}} sin ( θ + 3 π 2 ) = − cos θ cos ( θ + 3 π 2 ) = + sin θ tan ( θ + 3 π 2 ) = − cot θ cot ( θ + 3 π 2 ) = − tan θ sec ( θ + 3 π 2 ) = + csc θ csc ( θ + 3 π 2 ) = − sec θ {\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {3\pi }{2}})&=-\cos \theta \\\cos(\theta +{\tfrac {3\pi }{2}})&=+\sin \theta \\\tan(\theta +{\tfrac {3\pi }{2}})&=-\cot \theta \\\cot(\theta +{\tfrac {3\pi }{2}})&=-\tan \theta \\\sec(\theta +{\tfrac {3\pi }{2}})&=+\csc \theta \\\csc(\theta +{\tfrac {3\pi }{2}})&=-\sec \theta \end{aligned}}} sin ( θ + 2 π ) = + sin θ cos ( θ + 2 π ) = + cos θ tan ( θ + 2 π ) = + tan θ cot ( θ + 2 π ) = + cot θ sec ( θ + 2 π ) = + sec θ csc ( θ + 2 π ) = + csc θ {\displaystyle {\begin{aligned}\sin(\theta +2\pi )&=+\sin \theta \\\cos(\theta +2\pi )&=+\cos \theta \\\tan(\theta +2\pi )&=+\tan \theta \\\cot(\theta +2\pi )&=+\cot \theta \\\sec(\theta +2\pi )&=+\sec \theta \\\csc(\theta +2\pi )&=+\csc \theta \end{aligned}}} 關閉 角的和差恆等式 正弦與餘弦的角和公式的圖形證明法。使用了相似三角形的性質與三角函數的定義，強調的線段是單位長度 正切的角和公式的圖形證明法。使用了相似三角形的性質與三角函數的定義，強調的線段是單位長度。 又稱做「和差定理」、「和差公式」或「和角公式」。最簡要的檢定方式是使用歐拉公式[註 1]。 更多資訊 , ... 正弦 sin ( α ± β ) = sin α cos β ± cos α sin β {\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \,} 餘弦 cos ( α ± β ) = cos α cos β ∓ sin α sin β {\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \,} 正切 tan ( α ± β ) = tan α ± tan β 1 ∓ tan α tan β {\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}} 餘切 cot ( α ± β ) = cot α cot β ∓ 1 cot β ± cot α {\displaystyle \cot(\alpha \pm \beta )={\frac {\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }}} 正割 sec ( α ± β ) = sec α sec β 1 ∓ tan α tan β {\displaystyle \sec(\alpha \pm \beta )={\frac {\sec \alpha \sec \beta }{1\mp \tan \alpha \tan \beta }}} 餘割 csc ( α ± β ) = csc α csc β cot β ± cot α {\displaystyle \csc(\alpha \pm \beta )={\frac {\csc \alpha \csc \beta }{\cot \beta \pm \cot \alpha }}} 注意正負號的對應。 x ± y = a ± b ⇒ x + y = a + b and x − y = a − b {\displaystyle {\begin{aligned}x\pm y=a\pm b&\Rightarrow \ x+y=a+b\\&{\mbox{and}}\ x-y=a-b\end{aligned}}} x ± y = a ∓ b ⇒ x + y = a − b and x − y = a + b {\displaystyle {\begin{aligned}x\pm y=a\mp b&\Rightarrow \ x+y=a-b\\&{\mbox{and}}\ x-y=a+b\end{aligned}}} 關閉 根據 s i n π 4 = c o s π 4 = 1 2 {\displaystyle sin{\frac {\pi }{4}}=cos{\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}} ，以及和差恆等式，可以得到同角的正弦餘弦的和差關係，例如， sin α + cos α = 2 ( sin α cos π 4 + sin π 4 cos α ) = 2 sin ( α + π 4 ) = 2 cos ( α − π 4 ) {\displaystyle \sin \alpha +\cos \alpha ={\sqrt {2}}\left(\sin \alpha \cos {\frac {\pi }{4}}+\sin {\frac {\pi }{4}}\cos \alpha \right)={\sqrt {2}}\sin \left(\alpha +{\frac {\pi }{4}}\right)={\sqrt {2}}\cos \left(\alpha -{\frac {\pi }{4}}\right)} sin α − cos α = 2 ( sin α cos π 4 − sin π 4 cos α ) = 2 sin ( α − π 4 ) = 2 cos ( α + π 4 ) {\displaystyle \sin \alpha -\cos \alpha ={\sqrt {2}}\left(\sin \alpha \cos {\frac {\pi }{4}}-\sin {\frac {\pi }{4}}\cos \alpha \right)={\sqrt {2}}\sin \left(\alpha -{\frac {\pi }{4}}\right)={\sqrt {2}}\cos \left(\alpha +{\frac {\pi }{4}}\right)} 正弦與餘弦的無限多項和 sin ( ∑ i = 1 ∞ θ i ) = ∑ o d d k ≥ 1 ( − 1 ) k − 1 2 ∑ | A | = k ( ∏ i ∈ A sin θ i ∏ i ∉ A cos θ i ) {\displaystyle \sin \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {odd} \ k\geq 1}(-1)^{\frac {k-1}{2}}\sum _{|A|=k}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)} cos ( ∑ i = 1 ∞ θ i ) = ∑ e v e n k ≥ 0 ( − 1 ) k 2 ∑ | A | = k ( ∏ i ∈ A sin θ i ∏ i ∉ A cos θ i ) {\displaystyle \cos \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {even} \ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum _{|A|=k}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)} 這里的" | A | = k {\displaystyle |A|=k} "意味著索引 A {\displaystyle A} 遍歷集合 { 1 , 2 , 3 , … } {\displaystyle \left\{1,2,3,\ldots \right\}} 的大小為 k {\displaystyle k} 的所有子集的集合。 在這兩個恆等式中出現了在有限多項中不出現的不對稱：在每個乘積中，只有有限多個正弦因子和餘有限多個餘弦因子。 如果只有有限多項 θ i {\displaystyle \theta _{i}} 是非零，則在右邊只有有限多項是非零，因為正弦因子將變為零，而在每個項中，所有卻有限多的餘弦因子將是單位一。 正切的有限多項和 設 x i = tan θ i {\displaystyle x_{i}=\tan \theta _{i}} ，對於 i = 1 , … , n {\displaystyle i=1,\ldots ,n} 。設 e k {\displaystyle e_{k}} 是變量 x i {\displaystyle x_{i}} ， i = 1 , … , n {\displaystyle i=1,\ldots ,n} ， k = 0 , … , n {\displaystyle k=0,\ldots ,n} 的 k {\displaystyle k} 次基本對稱多項式。則 tan ( θ 1 + ⋯ + θ n ) = e 1 − e 3 + e 5 − ⋯ e 0 − e 2 + e 4 − ⋯ , {\displaystyle \tan(\theta _{1}+\cdots +\theta _{n})={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }},} 項的數目依賴於 n {\displaystyle n} 。例如， tan ( θ 1 + θ 2 + θ 3 ) = e 1 − e 3 e 0 − e 2 = ( x 1 + x 2 + x 3 ) − ( x 1 x 2 x 3 ) 1 − ( x 1 x 2 + x 1 x 3 + x 2 x 3 ) , tan ( θ 1 + θ 2 + θ 3 + θ 4 ) = e 1 − e 3 e 0 − e 2 + e 4 = ( x 1 + x 2 + x 3 + x 4 ) − ( x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 2 x 3 x 4 ) 1 − ( x 1 x 2 + x 1 x 3 + x 1 x 4 + x 2 x 3 + x 2 x 4 + x 3 x 4 ) + ( x 1 x 2 x 3 x 4 ) , {\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2}+\theta _{3})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\\\\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\\\&{}={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}} 並以此類推。一般情況可通過數學歸納法證明。 多倍角公式 更多資訊 , 是 ... T n {\displaystyle T_{n}} 是 n {\displaystyle n} 次切比雪夫多項式 cos n θ = T n cos θ {\displaystyle \cos n\theta =T_{n}\cos \theta \,} S n {\displaystyle S_{n}} 是 n {\displaystyle n} 次伸展多項式 sin 2 n θ = S n sin 2 θ {\displaystyle \sin ^{2}n\theta =S_{n}\sin ^{2}\theta \,} 棣莫弗定理， i {\displaystyle i} 是虛單位 cos n θ + i sin n θ = ( cos θ + i sin θ ) n {\displaystyle \cos n\theta +i\sin n\theta =(\cos \theta +i\sin \theta )^{n}\,} 關閉 1 + 2 cos x + 2 cos 2 x + 2 cos 3 x + ⋯ + 2 cos ( n x ) = sin [ ( n + 1 2 ) x ] sin x 2 {\displaystyle 1+2\cos x+2\cos 2x+2\cos 3x+\cdots +2\cos(nx)={\frac {\sin \left[\left(n+{\frac {1}{2}}\right)x\right]}{\sin {\frac {x}{2}}}}} 。 （這個 x {\displaystyle x} 的函數是狄利克雷核。） 雙倍角、三倍角和半角公式 這些公式可以使用和差恆等式或多倍角公式來證明。 更多資訊 , ... 弦 切 割 雙倍角公式 正 sin 2 θ = 2 sin θ cos θ = 2 tan θ 1 + tan 2 θ {\displaystyle {\begin{aligned}\sin 2\theta &=2\sin \theta \cos \theta \ \\&={\frac {2\tan \theta }{1+\tan ^{2}\theta }}\end{aligned}}} tan 2 θ = 2 tan θ 1 − tan 2 θ = 1 1 − tan θ − 1 1 + tan θ {\displaystyle {\begin{aligned}\tan 2\theta &={\frac {2\tan \theta }{1-\tan ^{2}\theta }}\ \\&={\frac {1}{1-\tan \theta }}-{\frac {1}{1+\tan \theta }}\end{aligned}}} sec 2 θ = sec 2 θ 1 − tan 2 θ = sec 2 θ 2 − sec 2 θ {\displaystyle {\begin{aligned}\sec 2\theta &={\frac {\sec ^{2}\theta }{1-\tan ^{2}\theta }}\\&={\frac {\sec ^{2}\theta }{2-\sec ^{2}\theta }}\end{aligned}}} 餘 cos 2 θ = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ = 1 − tan 2 θ 1 + tan 2 θ {\displaystyle {\begin{aligned}\cos 2\theta &=\cos ^{2}\theta -\sin ^{2}\theta \\&=2\cos ^{2}\theta -1\\&=1-2\sin ^{2}\theta \\&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\end{aligned}}} cot 2 θ = cot 2 θ − 1 2 cot θ = cot θ − tan θ 2 {\displaystyle {\begin{aligned}\cot 2\theta &={\frac {\cot ^{2}\theta -1}{2\cot \theta }}\\&={\frac {\cot \theta -\tan \theta }{2}}\end{aligned}}} csc 2 θ = csc 2 θ 2 cot θ = sec θ csc θ 2 {\displaystyle {\begin{aligned}\csc 2\theta &={\frac {\csc ^{2}\theta }{2\cot \theta }}\\&={\frac {\sec \theta \csc \theta }{2}}\end{aligned}}} 降次公式 正 sin 2 θ = 1 − cos 2 θ 2 {\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}} tan 2 θ = 1 − cos 2 θ 1 + cos 2 θ {\displaystyle \tan ^{2}\theta ={\frac {1-\cos 2\theta }{1+\cos 2\theta }}} 餘 cos 2 θ = 1 + cos 2 θ 2 {\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}} cot 2 θ = 1 + cos 2 θ 1 − cos 2 θ {\displaystyle \cot ^{2}\theta ={\frac {1+\cos 2\theta }{1-\cos 2\theta }}} 三倍角公式 正 sin 3 θ = 3 sin θ − 4 sin 3 θ = 4 sin θ sin ( π 3 − θ ) sin ( π 3 + θ ) {\displaystyle {\begin{aligned}\sin 3\theta &=3\sin \theta -4\sin ^{3}\theta \\&=4\sin \theta \sin \left({\frac {\pi }{3}}-\theta \right)\sin \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}} tan 3 θ = 3 tan θ − tan 3 θ 1 − 3 tan 2 θ = tan θ tan ( π 3 − θ ) tan ( π 3 + θ ) {\displaystyle {\begin{aligned}\tan 3\theta &={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}\\&=\tan \theta \tan \left({\frac {\pi }{3}}-\theta \right)\tan \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}} sec 3 θ = sec 3 θ 4 − 3 sec 2 θ = 1 4 cos θ cos ( π 3 − θ ) cos ( π 3 + θ ) {\displaystyle {\begin{aligned}\sec 3\theta &={\frac {\sec ^{3}\theta }{4-3\sec ^{2}\theta }}\\&={\dfrac {1}{4\cos \theta \cos \left({\dfrac {\pi }{3}}-\theta \right)\cos \left({\dfrac {\pi }{3}}+\theta \right)}}\end{aligned}}} 餘 cos 3 θ = 4 cos 3 θ − 3 cos θ = 4 cos θ cos ( π 3 − θ ) cos ( π 3 + θ ) {\displaystyle {\begin{aligned}\cos 3\theta &=4\cos ^{3}\theta -3\cos \theta \\&=4\cos \theta \cos \left({\frac {\pi }{3}}-\theta \right)\cos \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}} cot 3 θ = cot 3 θ − 3 cot θ 3 cot 2 θ − 1 = cot θ cot ( π 3 − θ ) cot ( π 3 + θ ) {\displaystyle {\begin{aligned}\cot 3\theta &={\frac {\cot ^{3}\theta -3\cot \theta }{3\cot ^{2}\theta -1}}\\&=\cot \theta \cot \left({\frac {\pi }{3}}-\theta \right)\cot \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}} csc 3 θ = csc 3 θ 3 csc 2 θ − 4 = 1 4 sin θ sin ( π 3 − θ ) sin ( π 3 + θ ) {\displaystyle {\begin{aligned}\csc 3\theta &={\frac {\csc ^{3}\theta }{3\csc ^{2}\theta -4}}\\&={\dfrac {1}{4\sin \theta \sin \left({\dfrac {\pi }{3}}-\theta \right)\sin \left({\dfrac {\pi }{3}}+\theta \right)}}\end{aligned}}} 半角公式 正 sin θ 2 = ± 1 − cos θ 2 {\displaystyle \sin {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{2}}}} tan θ 2 = csc θ − cot θ = ± 1 − cos θ 1 + cos θ = sin θ 1 + cos θ = 1 − cos θ sin θ = cos θ + sin θ − 1 cos θ − sin θ + 1 = cot 2 θ + 1 − cot θ {\displaystyle {\begin{aligned}\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta \\&=\pm \,{\sqrt {1-\cos \theta \over 1+\cos \theta }}\\&={\frac {\sin \theta }{1+\cos \theta }}\\&={\frac {1-\cos \theta }{\sin \theta }}\\&={\frac {\cos \theta +\sin \theta -1}{\cos \theta -\sin \theta +1}}\\&={\sqrt {\cot ^{2}\theta +1}}-\cot \theta \end{aligned}}} sec θ 2 = ± 2 sec θ sec θ + 1 {\displaystyle \sec {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {2\sec \theta }{\sec \theta +1}}}} 餘 cos θ 2 = ± 1 + cos θ 2 {\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}}} cot θ 2 = csc θ + cot θ = ± 1 + cos θ 1 − cos θ = sin θ 1 − cos θ = 1 + cos θ sin θ = cos θ − sin θ + 1 cos θ + sin θ − 1 = cot 2 θ + 1 + cot θ {\displaystyle {\begin{aligned}\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta \\&=\pm \,{\sqrt {1+\cos \theta \over 1-\cos \theta }}\\&={\frac {\sin \theta }{1-\cos \theta }}\\&={\frac {1+\cos \theta }{\sin \theta }}\\&={\frac {\cos \theta -\sin \theta +1}{\cos \theta +\sin \theta -1}}\\&={\sqrt {\cot ^{2}\theta +1}}+\cot \theta \end{aligned}}} csc θ 2 = ± 2 sec θ sec θ − 1 {\displaystyle \csc {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {2\sec \theta }{\sec \theta -1}}}} 關閉 n倍角公式 更多資訊 , ... n {\displaystyle n} 倍角公式 sin n θ = ∑ k = 0 n ( n k ) cos k θ sin n − k θ sin [ 1 2 ( n − k ) π ] = sin θ ∑ k = 0 ⌊ n − 1 2 ⌋ ( − 1 ) k ( n − 1 − k k ) ( 2 cos θ ) n − 1 − 2 k {\displaystyle \sin n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\sin \left[{\frac {1}{2}}(n-k)\pi \right]=\sin \theta \sum _{k=0}^{\lfloor {\frac {n-1}{2}}\rfloor }(-1)^{k}{\binom {n-1-k}{k}}~(2\cos \theta )^{n-1-2k}} （第二類切比雪夫多項式） cos n θ = ∑ k = 0 n ( n k ) cos k θ sin n − k θ cos [ 1 2 ( n − k ) π ] = 1 2 ∑ k = 0 ⌊ n 2 ⌋ ( − 1 ) k n n − k ( n − k k ) ( 2 cos θ ) n − 2 k {\displaystyle \cos n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\cos \left[{\frac {1}{2}}(n-k)\pi \right]={\frac {1}{2}}\sum _{k=0}^{\lfloor {\frac {n}{2}}\rfloor }(-1)^{k}{\frac {n}{n-k}}{\binom {n-k}{k}}~(2\cos \theta )^{n-2k}} （第一類切比雪夫多項式） tan n θ = ∑ k = 1 [ n 2 ] ( − 1 ) k + 1 ( n 2 k − 1 ) tan 2 k − 1 θ ∑ k = 1 [ n + 1 2 ] ( − 1 ) k + 1 ( n 2 ( k − 1 ) ) tan 2 ( k − 1 ) θ {\displaystyle \tan n\theta ={\frac {\displaystyle \sum _{k=1}^{\left[{\frac {n}{2}}\right]}(-1)^{k+1}{\binom {n}{2k-1}}\tan ^{2k-1}\theta }{\displaystyle \sum _{k=1}^{\left[{\frac {n+1}{2}}\right]}(-1)^{k+1}{\binom {n}{2(k-1)}}\tan ^{2(k-1)}\theta }}} n {\displaystyle n} 倍遞迴公式 tan n θ = tan ( n − 1 ) θ + tan θ 1 − tan ( n − 1 ) θ tan θ {\displaystyle \tan \,n\theta ={\frac {\tan(n{-}1)\theta +\tan \theta }{1-\tan(n{-}1)\theta \,\tan \theta }}} 、 cot n θ = cot ( n − 1 ) θ cot θ − 1 cot ( n − 1 ) θ + cot θ {\displaystyle \cot \,n\theta ={\frac {\cot(n{-}1)\theta \,\cot \theta -1}{\cot(n{-}1)\theta +\cot \theta }}} 。(遞迴關係) 關閉 參見正切半角公式，它也叫做「萬能公式」。 其他函數的倍半角公式 正矢 versin 2 θ = 2 sin 2 θ = ( sin 2 θ ) ( sin θ ) cos θ = 1 − cos 2 θ {\displaystyle \operatorname {versin} 2\theta =2\sin ^{2}\theta ={\frac {(\sin 2\theta )(\sin \theta )}{\cos \theta }}=1-\cos 2\theta } 餘矢 cvs 2 θ = ( sin θ − cos θ ) 2 = 1 − sin 2 θ {\displaystyle \operatorname {cvs} 2\theta =(\sin \theta -\cos \theta )^{2}=1-\sin 2\theta } 冪簡約公式 從解餘弦二倍角公式的第二和第三版本得到。 更多資訊 , ... 正弦 餘弦 其他 sin 2 θ = 1 − cos 2 θ 2 {\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}} cos 2 θ = 1 + cos 2 θ 2 {\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}} sin 2 θ cos 2 θ = 1 − cos 4 θ 8 {\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos 4\theta }{8}}} sin 3 θ = 3 sin θ − sin 3 θ 4 {\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin 3\theta }{4}}} cos 3 θ = 3 cos θ + cos 3 θ 4 {\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos 3\theta }{4}}} sin 3 θ cos 3 θ = 3 sin 2 θ − sin 6 θ 32 {\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin 2\theta -\sin 6\theta }{32}}} sin 4 θ = 3 − 4 cos 2 θ + cos 4 θ 8 {\displaystyle \sin ^{4}\theta ={\frac {3-4\cos 2\theta +\cos 4\theta }{8}}} cos 4 θ = 3 + 4 cos 2 θ + cos 4 θ 8 {\displaystyle \cos ^{4}\theta ={\frac {3+4\cos 2\theta +\cos 4\theta }{8}}} sin 4 θ cos 4 θ = 3 − 4 cos 4 θ + cos 8 θ 128 {\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos 4\theta +\cos 8\theta }{128}}} sin 5 θ = 10 sin θ − 5 sin 3 θ + sin 5 θ 16 {\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin 3\theta +\sin 5\theta }{16}}} cos 5 θ = 10 cos θ + 5 cos 3 θ + cos 5 θ 16 {\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos 3\theta +\cos 5\theta }{16}}} sin 5 θ cos 5 θ = 10 sin 2 θ − 5 sin 6 θ + sin 10 θ 512 {\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin 2\theta -5\sin 6\theta +\sin 10\theta }{512}}} 關閉 更多資訊 如果 ... 餘弦 正弦 如果 n {\displaystyle n} 是奇數 cos n θ = 2 2 n ∑ k = 0 n − 1 2 ( n k ) cos [ ( n − 2 k ) θ ] {\displaystyle \cos ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}{\binom {n}{k}}\cos {[(n-2k)\theta ]}} sin n θ = 2 2 n ∑ k = 0 n − 1 2 ( − 1 ) ( n − 1 2 − k ) ( n k ) sin [ ( n − 2 k ) θ ] {\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}(-1)^{\left({\frac {n-1}{2}}-k\right)}{\binom {n}{k}}\sin {[(n-2k)\theta ]}} 如果 n {\displaystyle n} 是偶數 cos n θ = 1 2 n ( n n 2 ) + 2 2 n ∑ k = 0 n 2 − 1 ( n k ) cos [ ( n − 2 k ) θ ] {\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {[(n-2k)\theta ]}} sin n θ = 1 2 n ( n n 2 ) + 2 2 n ∑ k = 0 n 2 − 1 ( − 1 ) ( n 2 − k ) ( n k ) cos [ ( n − 2 k ) θ ] {\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{\left({\frac {n}{2}}-k\right)}{\binom {n}{k}}\cos {[(n-2k)\theta ]}} 關閉 數值連乘 ∏ k = 0 n − 1 cos 2 k θ = sin 2 n θ 2 n sin θ {\displaystyle \prod _{k=0}^{n-1}\cos 2^{k}\theta ={\frac {\sin 2^{n}\theta }{2^{n}\sin \theta }}} [2] ∏ k = 0 n − 1 sin ( x + k π n ) = sin n x 2 n − 1 {\displaystyle \prod _{k=0}^{n-1}\sin \left(x+{\frac {k\pi }{n}}\right)={\frac {\sin nx}{2^{n-1}}}} [2] ∏ k = 1 n − 1 sin ( k π n ) = n 2 n − 1 {\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}} , ∏ k = 1 n − 1 sin ( k π 2 n ) = n 2 n − 1 {\displaystyle \prod _{k=1}^{n-1}\sin \left({\frac {k\pi }{2n}}\right)={\frac {\sqrt {n}}{2^{n-1}}}} , ∏ k = 1 n sin ( k π 2 n + 1 ) = 2 n + 1 2 n {\displaystyle \prod _{k=1}^{n}\sin \left({\frac {k\pi }{2n+1}}\right)={\frac {\sqrt {2n+1}}{2^{n}}}} ∏ k = 1 n − 1 cos ( k π n ) = sin n π 2 2 n − 1 {\displaystyle \prod _{k=1}^{n-1}\cos \left({\frac {k\pi }{n}}\right)={\frac {\sin {\frac {n\pi }{2}}}{2^{n-1}}}} , ∏ k = 1 n − 1 cos ( k π 2 n ) = n 2 n − 1 {\displaystyle \prod _{k=1}^{n-1}\cos \left({\frac {k\pi }{2n}}\right)={\frac {\sqrt {n}}{2^{n-1}}}} , ∏ k = 1 n cos ( k π 2 n + 1 ) = 1 2 n {\displaystyle \prod _{k=1}^{n}\cos \left({\frac {k\pi }{2n+1}}\right)={\frac {1}{2^{n}}}} ∏ k = 1 n − 1 tan ( k π n ) = n sin n π 2 {\displaystyle \prod _{k=1}^{n-1}\tan \left({\frac {k\pi }{n}}\right)={\frac {n}{\sin {\frac {n\pi }{2}}}}} , ∏ k = 1 n − 1 tan ( k π 2 n ) = 1 {\displaystyle \prod _{k=1}^{n-1}\tan \left({\frac {k\pi }{2n}}\right)=1} , ∏ k = 1 n tan k π 2 n + 1 = 2 n + 1 {\displaystyle \prod _{k=1}^{n}\tan {\frac {k\pi }{2n+1}}={\sqrt {2n+1}}} 常見的恆等式 積化和差與和差化積恆等式 數學家韋達在其三角學著作《應用於三角形的數學定律》給出積化和差與和差化積恆等式。積化和差恆等式可以通過展開角的和差恆等式的右手端來證明。 更多資訊 , ... 積化和差 和差化積 sin α cos β = sin ( α + β ) + sin ( α − β ) 2 {\displaystyle \sin \alpha \cos \beta ={\sin(\alpha +\beta )+\sin(\alpha -\beta ) \over 2}} sin α + sin β = 2 sin α + β 2 cos α − β 2 {\displaystyle \sin \alpha +\sin \beta =2\sin {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}} cos α sin β = sin ( α + β ) − sin ( α − β ) 2 {\displaystyle \cos \alpha \sin \beta ={\sin(\alpha +\beta )-\sin(\alpha -\beta ) \over 2}} sin α − sin β = 2 cos α + β 2 sin α − β 2 {\displaystyle \sin \alpha -\sin \beta =2\cos {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2}} cos α cos β = cos ( α + β ) + cos ( α − β ) 2 {\displaystyle \cos \alpha \cos \beta ={\cos(\alpha +\beta )+\cos(\alpha -\beta ) \over 2}} cos α + cos β = 2 cos α + β 2 cos α − β 2 {\displaystyle \cos \alpha +\cos \beta =2\cos {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}} sin α sin β = − cos ( α + β ) − cos ( α − β ) 2 {\displaystyle \sin \alpha \sin \beta =-{\cos(\alpha +\beta )-\cos(\alpha -\beta ) \over 2}} cos α − cos β = − 2 sin α + β 2 sin α − β 2 {\displaystyle \cos \alpha -\cos \beta =-2\sin {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2}} 關閉 平方差公式 sin ( x + y ) sin ( x − y ) = sin 2 x − sin 2 y = cos 2 y − cos 2 x {\displaystyle \sin(x+y)\sin(x-y)=\sin ^{2}{x}-\sin ^{2}{y}=\cos ^{2}{y}-\cos ^{2}{x}\,} cos ( x + y ) cos ( x − y ) = cos 2 x − sin 2 y = cos 2 y − sin 2 x {\displaystyle \cos(x+y)\cos(x-y)=\cos ^{2}{x}-\sin ^{2}{y}=\cos ^{2}{y}-\sin ^{2}{x}\,} (可藉由積化和差公式+2倍角公式推導而來) 其他恆等式 如果 x + y + z = n π {\displaystyle x+y+z=n\pi } ， 那麼 tan x + tan y + tan z = tan x tan y tan z {\displaystyle \tan x+\tan y+\tan z=\tan x\tan y\tan z} cot x cot y + cot y cot z + cot z cot x = 1 {\displaystyle \cot x\cot y+\cot y\cot z+\cot z\cot x=1} 如果 x + y + z = n π + π 2 {\displaystyle x+y+z=n\pi +{\frac {\pi }{2}}} ， 那麼 tan x tan y + tan y tan z + tan z tan x = 1 {\displaystyle \tan x\tan y+\tan y\tan z+\tan z\tan x=1} cot x + cot y + cot z = cot x cot y cot z {\displaystyle \cot x+\cot y+\cot z=\cot x\cot y\cot z} 如果 x + y + z = π {\displaystyle x+y+z=\pi } ， 那麼 sin 2 x + sin 2 y + sin 2 z = 4 sin x sin y sin z {\displaystyle \sin 2x+\sin 2y+\sin 2z=4\sin x\sin y\sin z} sin x + sin y + sin z = 4 cos x 2 cos y 2 cos z 2 {\displaystyle \sin x+\sin y+\sin z=4\cos {\frac {x}{2}}\cos {\frac {y}{2}}\cos {\frac {z}{2}}} cos x + cos y + cos z = 1 + 4 sin x 2 sin y 2 sin z 2 {\displaystyle \cos x+\cos y+\cos z=1+4\sin {\frac {x}{2}}\sin {\frac {y}{2}}\sin {\frac {z}{2}}} 托勒密定理 如果 w + x + y + z = π {\displaystyle w+x+y+z=\pi } （半圓） 那麼: sin ( w + x ) sin ( x + y ) = sin ( x + y ) sin ( y + z ) = sin ( y + z ) sin ( z + w ) = sin ( z + w ) sin ( w + x ) = sin w sin y + sin x sin z {\displaystyle {\begin{aligned}\sin(w+x)\sin(x+y)&=\sin(x+y)\sin(y+z)\\&=\sin(y+z)\sin(z+w)\\&{}=\sin(z+w)\sin(w+x)\\&{}=\sin w\sin y+\sin x\sin z\end{aligned}}} （前三個等式是一般情況；第四個是本質。） 三角函數與雙曲函數的恆等式 利用三角恆等式的指數定義和雙曲函數的指數定義即可求出下列恆等式： e i x = cos x + i sin x , e − i x = cos x − i sin x {\displaystyle e^{ix}=\cos x+i\;\sin x,\;e^{-ix}=\cos x-i\;\sin x} e x = cosh x + sinh x , e − x = cosh x − sinh x {\displaystyle e^{x}=\cosh x+\sinh x\!,\;e^{-x}=\cosh x-\sinh x\!} 所以 cosh i x = 1 2 ( e i x + e − i x ) = cos x {\displaystyle \cosh ix={\tfrac {1}{2}}(e^{ix}+e^{-ix})=\cos x} sinh i x = 1 2 ( e i x − e − i x ) = i sin x {\displaystyle \sinh ix={\tfrac {1}{2}}(e^{ix}-e^{-ix})=i\sin x} 下表列出部分的三角函數與雙曲函數的恆等式： 更多資訊 , ... 三角函數 雙曲函數 sin θ = − i sinh i θ {\displaystyle \sin \theta =-i\sinh {i\theta }\,} sinh θ = i sin ( − i θ ) {\displaystyle \sinh {\theta }=i\sin {(-i\theta )}\,} cos θ = cosh i θ {\displaystyle \cos {\theta }=\cosh {i\theta }\,} cosh θ = cos ( − i θ ) {\displaystyle \cosh {\theta }=\cos {(-i\theta )}\,} tan θ = − i tanh i θ {\displaystyle \tan \theta =-i\tanh {i\theta }\,} tanh θ = i tan ( − i θ ) {\displaystyle \tanh {\theta }=i\tan {(-i\theta )}\,} cot θ = i coth i θ {\displaystyle \cot {\theta }=i\coth {i\theta }\,} coth θ = − i cot ( − i θ ) {\displaystyle \coth \theta =-i\cot {(-i\theta )}\,} sec θ = sech i θ {\displaystyle \sec {\theta }=\operatorname {sech} {\,i\theta }\,} sech θ = sec ( − i θ ) {\displaystyle \operatorname {sech} {\theta }=\sec {(-i\theta )}\,} csc θ = i csch i θ {\displaystyle \csc {\theta }=i\;\operatorname {csch} {\,i\theta }\,} csch θ = − i csc ( − i θ ) {\displaystyle \operatorname {csch} \theta =-i\csc {(-i\theta )}\,} 關閉 其他恆等式： cosh i x = 1 2 ( e i x + e − i x ) = cos x {\displaystyle \cosh ix={\tfrac {1}{2}}(e^{ix}+e^{-ix})=\cos x} sinh i x = 1 2 ( e i x − e − i x ) = i sin x {\displaystyle \sinh ix={\tfrac {1}{2}}(e^{ix}-e^{-ix})=i\sin x} cosh ( x + i y ) = cosh ( x ) cos ( y ) + i sinh ( x ) sin ( y ) {\displaystyle \cosh(x+iy)=\cosh(x)\cos(y)+i\sinh(x)\sin(y)\,} sinh ( x + i y ) = sinh ( x ) cos ( y ) + i cosh ( x ) sin ( y ) {\displaystyle \sinh(x+iy)=\sinh(x)\cos(y)+i\cosh(x)\sin(y)\,} tanh i x = i tan x {\displaystyle \tanh ix=i\tan x\,} cosh x = cos i x {\displaystyle \cosh x=\cos ix\,} sinh x = − i sin i x {\displaystyle \sinh x=-i\sin ix\,} tanh x = − i tan i x {\displaystyle \tanh x=-i\tan ix\,} 線性組合 對於某些用途，知道同樣周期但不同相位移動的正弦波的任何線性組合是有相同周期但不同相位移動的正弦波是重要的。在正弦和餘弦波的線性組合的情況下，我們有 a sin x + b cos x = a 2 + b 2 ⋅ sin ( x + φ ) ( a > 0 ) {\displaystyle a\sin x+b\cos x={\sqrt {a^{2}+b^{2}}}\cdot \sin(x+\varphi )\;(a>0)} 這裡的 φ = arctan ( b a ) {\displaystyle \varphi =\arctan \left({\frac {b}{a}}\right)} 這個公式也叫輔助角公式或李善蘭公式。更一般的說，對於任何相位移動，我們有 a sin x + b sin ( x + α ) = c sin ( x + β ) ( a + b cos x > 0 ) {\displaystyle a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\;(a+b\cos x>0)} 這裡 c = a 2 + b 2 + 2 a b cos α , {\displaystyle c={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }},} 而 β = arctan ( b sin α a + b cos α ) {\displaystyle \beta =\arctan \left({\frac {b\sin \alpha }{a+b\cos \alpha }}\right)} 反三角函數 主條目：反三角函數 arcsin x + arccos x = π 2 {\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}}\;} arctan x + arccot x = π 2 . {\displaystyle \arctan x+\operatorname {arccot} x={\frac {\pi }{2}}.\;} arctan x + arctan 1 x = { π 2 , if x > 0 − π 2 , if x < 0 {\displaystyle \arctan x+\arctan {\frac {1}{x}}=\left\{{\begin{matrix}{\frac {\pi }{2}},&{\mbox{if }}x>0\\-{\frac {\pi }{2}},&{\mbox{if }}x<0\end{matrix}}\right.} arctan x + arctan y = arctan x + y 1 − x y + { π , if x , y > 0 − π , if x , y < 0 0 , otherwise {\displaystyle \arctan x+\arctan y=\arctan {\frac {x+y}{1-xy}}+\left\{{\begin{matrix}\pi ,&{\mbox{if }}x,y>0\\-\pi ,&{\mbox{if }}x,y<0\\0,&{\mbox{otherwise }}\end{matrix}}\right.} sin ( arccos x ) = 1 − x 2 {\displaystyle \sin(\arccos x)={\sqrt {1-x^{2}}}\,} sin ( arctan x ) = x 1 + x 2 {\displaystyle \sin(\arctan x)={\frac {x}{\sqrt {1+x^{2}}}}} cos ( arctan x ) = 1 1 + x 2 {\displaystyle \cos(\arctan x)={\frac {1}{\sqrt {1+x^{2}}}}} cos ( arcsin x ) = 1 − x 2 {\displaystyle \cos(\arcsin x)={\sqrt {1-x^{2}}}\,} tan ( arcsin x ) = x 1 − x 2 {\displaystyle \tan(\arcsin x)={\frac {x}{\sqrt {1-x^{2}}}}} tan ( arccos x ) = 1 − x 2 x {\displaystyle \tan(\arccos x)={\frac {\sqrt {1-x^{2}}}{x}}} 無限乘積公式 為了用於特殊函數，有下列三角函數無窮乘積公式[3][4]： sin x = x ∏ n = 1 ∞ ( 1 − x 2 π 2 n 2 ) {\displaystyle \sin x=x\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)} sinh x = x ∏ n = 1 ∞ ( 1 + x 2 π 2 n 2 ) {\displaystyle \sinh x=x\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)} sin x x = ∏ n = 1 ∞ cos ( x 2 n ) {\displaystyle {\frac {\sin x}{x}}=\prod _{n=1}^{\infty }\cos \left({\frac {x}{2^{n}}}\right)} cos x = x ∏ n = 1 ∞ ( 1 − x 2 π 2 ( n − 1 2 ) 2 ) {\displaystyle \cos x=x\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}(n-{\frac {1}{2}})^{2}}}\right)} cosh x = ∏ n = 1 ∞ ( 1 + x 2 π 2 ( n − 1 2 ) 2 ) {\displaystyle \cosh x=\prod _{n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}(n-{\frac {1}{2}})^{2}}}\right)} | sin x | = 1 2 ∏ n = 0 ∞ | tan ( 2 n x ) | 2 n + 1 {\displaystyle |\sin x|={\frac {1}{2}}\prod _{n=0}^{\infty }{\sqrt[{2^{n+1}}]{\left|\tan \left(2^{n}x\right)\right|}}} 微積分 正弦的微分 由於已知的技術原因，圖表暫時不可用。帶來不便，我們深表歉意。 正弦（藍色）、正弦的微分（橘色），其中，正弦的微分正好是餘弦。 餘弦的微分 由於已知的技術原因，圖表暫時不可用。帶來不便，我們深表歉意。 餘弦（藍色）、餘弦的微分（橘色），其中，餘弦的微分正好是正弦的對x軸的鏡射。 在微積分中，下面陳述的關係要求角用弧度來度量；如果用其他方式比如角度來這些關係會變得更加複雜。如果三角函數以幾何的方式來定義，它們的導數可以通過驗證兩個極限而找到。第一個是： lim x → 0 sin x x = 1 {\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1} 可以使用單位圓和夾擠定理來驗證。如果用洛必達法則來證明這個極限，那也就用這個極限證明了正弦的導數是餘弦，並因此在應用洛必達法則中使用正弦的導數是餘弦的事實，就是邏輯謬論中的循環論證了。第二個極限是： lim x → 0 cos x − 1 x = 0 {\displaystyle \lim _{x\rightarrow 0}{\frac {\cos x-1}{x}}=0} 使用恆等式 tan x 2 = 1 − cos x sin x {\displaystyle \tan {\frac {x}{2}}=1-{\frac {\cos x}{\sin x}}} 驗證。已經確立了這兩個極限，你可以使用導數的極限定義和加法定理來證明 sin ′ x = cos x {\displaystyle \sin 'x=\cos x} 和 cos ′ x = − sin x {\displaystyle \cos 'x=-\sin x} 。如果正弦和餘弦函數用它們的泰勒級數來定義，則導數可以通過冪級數逐項微分得到。 d d x sin ( x ) = cos ( x ) {\displaystyle {d \over dx}\sin(x)=\cos(x)} 結果的三角函數可以使用上述恆等式和微分規則來做微分。 d d x sin x = cos x , d d x arcsin x = 1 1 − x 2 d d x cos x = − sin x , d d x arccos x = − 1 1 − x 2 d d x tan x = sec 2 x , d d x arctan x = 1 1 + x 2 d d x cot x = − csc 2 x , d d x arccot x = − 1 1 + x 2 d d x sec x = tan x sec x , d d x arcsec x = 1 | x | x 2 − 1 d d x csc x = − csc x cot x , d d x arccsc x = − 1 | x | x 2 − 1 {\displaystyle {\begin{aligned}{d \over dx}\sin x&=\cos x,&{d \over dx}\arcsin x&={1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\cos x&=-\sin x,&{d \over dx}\arccos x&=-{1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\tan x&=\sec ^{2}x,&{d \over dx}\arctan x&={1 \over 1+x^{2}}\\\\{d \over dx}\cot x&=-\csc ^{2}x,&{d \over dx}\operatorname {arccot} x&=-{1 \over 1+x^{2}}\\\\{d \over dx}\sec x&=\tan x\sec x,&{d \over dx}\operatorname {arcsec} x&={1 \over |x|{\sqrt {x^{2}-1}}}\\\\{d \over dx}\csc x&=-\csc x\cot x,&{d \over dx}\operatorname {arccsc} x&=-{1 \over |x|{\sqrt {x^{2}-1}}}\end{aligned}}} 在三角函數積分表中可以找到積分恆等式。 蘊涵 三角函數（正弦和餘弦）的微分是同樣兩個函數線性組合的事實在很多數學領域包括微分方程和傅立葉變換中是重要的基本原理。 指數定義 更多資訊 , ... 函數 反函數 sin θ = e i θ − e − i θ 2 i {\displaystyle \sin \theta ={\frac {e^{{i}\theta }-e^{-{i}\theta }}{2{i}}}\,} arcsin x = − i ln ( i x + 1 − x 2 ) {\displaystyle \arcsin x=-{i}\ln \left({i}x+{\sqrt {1-x^{2}}}\right)\,} cos θ = e i θ + e − i θ 2 {\displaystyle \cos \theta ={\frac {e^{{i}\theta }+e^{-{i}\theta }}{2}}\,} arccos x = − i ln ( x + x 2 − 1 ) {\displaystyle \arccos x=-{i}\ln \left(x+{\sqrt {x^{2}-1}}\right)\,} tan θ = e i θ − e − i θ i ( e i θ + e − i θ ) {\displaystyle \tan \theta ={\frac {e^{{i}\theta }-e^{-{i}\theta }}{{i}(e^{{i}\theta }+e^{-{i}\theta })}}\,} arctan x = i 2 ln ( i + x i − x ) {\displaystyle \arctan x={\frac {i}{2}}\ln \left({\frac {{i}+x}{{i}-x}}\right)\,} csc θ = 2 i e i θ − e − i θ {\displaystyle \csc \theta ={\frac {2{i}}{e^{{i}\theta }-e^{-{i}\theta }}}\,} arccsc x = − i ln ( i x + 1 − 1 x 2 ) {\displaystyle \operatorname {arccsc} x=-{i}\ln \left({\tfrac {i}{x}}+{\sqrt {1-{\tfrac {1}{x^{2}}}}}\right)\,} sec θ = 2 e i θ + e − i θ {\displaystyle \sec \theta ={\frac {2}{e^{{i}\theta }+e^{-{i}\theta }}}\,} arcsec x = − i ln ( 1 x + 1 − i x 2 ) {\displaystyle \operatorname {arcsec} x=-{i}\ln \left({\tfrac {1}{x}}+{\sqrt {1-{\tfrac {i}{x^{2}}}}}\right)\,} cot θ = i ( e i θ + e − i θ ) e i θ − e − i θ {\displaystyle \cot \theta ={\frac {{i}(e^{{i}\theta }+e^{-{i}\theta })}{e^{{i}\theta }-e^{-{i}\theta }}}\,} arccot x = i 2 ln ( i − x i + x ) {\displaystyle \operatorname {arccot} x={\frac {i}{2}}\ln \left({\frac {{i}-x}{{i}+x}}\right)\,} cis θ = e i θ {\displaystyle \operatorname {cis} \,\theta =e^{{i}\theta }\,} arccis x = − i ln x {\displaystyle \operatorname {arccis} \,x=-{i}\ln x\,} sinh θ = e θ − e − θ 2 {\displaystyle \sinh \theta ={\frac {e^{\theta }-e^{-\theta }}{2}}\,} arcsinh x = ln ( x ± x 2 + 1 ) {\displaystyle \operatorname {arcsinh} \,x=\ln \left(x\pm {\sqrt {x^{2}+1}}\right)\,} cosh θ = e θ + e − θ 2 {\displaystyle \cosh \theta ={\frac {e^{\theta }+e^{-\theta }}{2}}\,} arccosh x = ln ( x ± x 2 − 1 ) = ± ln ( x + x 2 − 1 ) {\displaystyle \operatorname {arccosh} \,x=\ln \left(x\pm {\sqrt {x^{2}-1}}\right)=\pm \ln \left(x+{\sqrt {x^{2}-1}}\right)\,} tanh θ = sinh θ cosh θ = e θ − e − θ e θ + e − θ {\displaystyle \tanh \theta ={\frac {\sinh \theta }{\cosh \theta }}={\frac {e^{\theta }-e^{-\theta }}{e^{\theta }+e^{-\theta }}}\,} arctanh x = 1 2 ln ( 1 + x 1 − x ) {\displaystyle \operatorname {arctanh} \,x={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)\,} 關閉 參見 餘弦定理 三角函數 正弦定理 正切定理 中線長公式 角平分線長公式 雙曲函數恆等式 三分之一角公式 註釋 [註 1]由於歐拉公式的證明過程中使用了棣莫弗公式，而棣莫弗公式的證明過程中使用了和角公式，故使用歐拉公式證明和角公式會造成循環論證，故而此方法僅為檢定方法，而非嚴謹的證明方法。對於類似方法也應注意甄別。 參考文獻 引用 [1]Abramowitz and Stegun, p. 78, 4.3.147 [2]蘇學孟. 求三角函数乘积的常用方法. 中學數學教學. 1995, (6) [2014-12-27]. （原始內容存檔於2014-12-27）. [3]Abramowitz and Stegun, p. 75, 4.3.89–90 [4]Abramowitz and Stegun, p. 85, 4.5.68–69 來源 A one-page proof of many trigonometric identities using Euler's formula，by Connelly Barnes. Useful Formulae for Hong Kong Advanced Level Examination Pure Mathematics. Wikiwand - on Seamless Wikipedia browsing. On steroids.
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...![{\displaystyle \cos n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\cos \left[{\frac {1}{2}}(n-k)\pi \right]={\frac {1}{2}}\sum _{k=0}^{\lfloor {\frac {n}{2}}\rfloor }(-1)^{k}{\frac {n}{n-k}}{\binom {n-k}{k}}~(2\cos \theta )^{n-2k}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/4aab0b9f9ec4bc519564a1720f9d1e10338c0c2f)
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![{\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{\left({\frac {n}{2}}-k\right)}{\binom {n}{k}}\cos {[(n-2k)\theta ]}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/450f6c0089bc8c2f424338bfd77e2efddf1f2cef)
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