餘弦(cosine)是三角函數的一種。它的定義域是整個實數集,值域是 [ − 1 , 1 ] {\displaystyle [-1,1]} 。它是周期函數,其最小正周期為 2 π {\displaystyle 2\pi } (360°)。在自變量為 2 n π {\displaystyle 2n\pi } (或 360 ∘ n {\displaystyle 360^{\circ }n} ,其中 n {\displaystyle n} 為整數)時,該函數有極大值1;在自變量為 ( 2 n + 1 ) π {\displaystyle (2n+1)\pi } ( 360 ∘ n + 180 ∘ {\displaystyle 360^{\circ }n+180^{\circ }} )時,該函數有極小值-1。餘弦函數是偶函數,其圖像關於y軸對稱。 餘弦 性質 奇偶性 偶 定義域 (-∞,∞) 到達域 [-1,1] 周期 2 π {\displaystyle 2\pi } (360°) 特定值 當x=0 1 當x=+∞ N/A 當x=-∞ N/A 最大值 (2 k π {\displaystyle k\pi } , 1)(360°k, 1) 最小值 ( ( 2 k + 1 ) π {\displaystyle \left(2k+1\right)\pi } , -1)(360°k+180°, -1) 其他性質 漸近線 N/A 根 k π − π 2 {\displaystyle k\pi -{\tfrac {\pi }{2}}} ( 180 ∘ k − 90 ∘ {\displaystyle 180^{\circ }k-90^{\circ }} ) 臨界點 k π {\displaystyle k\pi } ( 180 ∘ k {\displaystyle 180^{\circ }k} ) 拐點 k π − π 2 {\displaystyle k\pi -{\tfrac {\pi }{2}}} ( 180 ∘ k − 90 ∘ {\displaystyle 180^{\circ }k-90^{\circ }} ) 不動點 x軸為弧度時: 0.7390851332152...(42.3464588340929...°) x軸為角度時:0.999847741531088...°(0.0174506351083467...) k是一個整數。 餘弦的符號為 cos {\displaystyle \cos } ,取自拉丁文cosinus。該符號最早由瑞士數學家萊昂哈德·歐拉所採用。 直角三角形中 直角三角形,∠C為直角,∠A 的角度為 θ {\displaystyle \theta } , 對於 ∠A 而言,a為對邊、b為鄰邊、c為斜邊 在直角三角形中,一個銳角 ∠ A {\displaystyle \angle A} 的餘弦定義為它的鄰邊與斜邊的比值,也就是: cos θ = b c {\displaystyle \cos \theta ={\frac {\mathrm {b} }{\mathrm {c} }}\,\!} 可以發現其定義和正割函數互為倒數。 直角坐標系中 設 α {\displaystyle \alpha } 是平面直角坐標系xOy中的一個象限角, P ( x , y ) {\displaystyle P\left({x,y}\right)} 是角的終邊上一點, r = x 2 + y 2 > 0 {\displaystyle r={\sqrt {x^{2}+y^{2}}}>0} 是P到原點O的距離,則 α {\displaystyle \alpha } 的餘弦定義為: cos α = x r {\displaystyle \cos \alpha ={\frac {x}{r}}\,\!} 單位圓定義 單位圓 圖像中給出了用弧度度量的某個公共角。逆時針方向的度量是正角,而順時針的度量是負角。設一個過原點的線,同x軸正半部分得到一個角 θ {\displaystyle \theta } ,並與單位圓相交。這個交點的y坐標等於 sin θ {\displaystyle \sin \theta } 。 在這個圖形中的三角形確保了這個公式;半徑等於斜邊並有長度1,所以有了 cos θ = x 1 {\displaystyle \cos \theta ={\frac {x}{1}}} 。單位圓可以被認為是通過改變鄰邊和對邊的長度並保持斜邊等於1查看無限數目的三角形的一種方式。 對於大於 2 π {\displaystyle 2\pi } (360°)或小於 − 2 π {\displaystyle -2\pi } (-360°)的角度,簡單的繼續繞單位圓旋轉。在這種方式下,餘弦變成了周期為 2 π {\displaystyle 2\pi } (360°)的周期函數: cos θ = cos ( θ + 2 π k ) = cos ( θ + 360 ∘ k ) {\displaystyle \cos \theta =\cos \left(\theta +2\pi k\right)=\cos \left(\theta +360^{\circ }k\right)} 對於任何角度 θ {\displaystyle \theta } 和任何整數 k {\displaystyle k} 。 級數定義 cos x = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! + ⋯ = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! {\displaystyle \cos x=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}\,\!} 微分方程定義 由於餘弦的導數是負的正弦,正弦的導數是餘弦,因此餘弦函數滿足初值問題 y ″ = − y , y ( 0 ) = 1 , y ′ ( 0 ) = 0 {\displaystyle y''=-y,\,y(0)=1,\,y'(0)=0} 這就是餘弦的微分方程定義。 指數定義 cos θ = e i θ + e − i θ 2 {\displaystyle \cos \theta ={\frac {e^{i\theta }+e^{-i\theta }}{2}}\,\!} 用其它三角函數來表示餘弦 More information , ... 函數 sin cos tan csc sec cot 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 }}}} csc 2 θ − 1 csc θ {\displaystyle {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}} 1 sec θ {\displaystyle {\frac {1}{\sec \theta }}} cot θ 1 + cot 2 θ {\displaystyle {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}} Close 兩角和差公式 cos ( x + y ) = cos x cos y − sin x sin y {\displaystyle \cos \left(x+y\right)=\cos x\cos y-\sin x\sin y} cos ( x − y ) = cos x cos y + sin x sin y {\displaystyle \cos \left(x-y\right)=\cos x\cos y+\sin x\sin y} 二倍角公式 cos ( 2 θ ) = cos 2 θ − sin 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ {\displaystyle \cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1=1-2\sin ^{2}\theta } 三倍角公式 cos 3 θ = 4 cos 3 θ − 3 cos θ {\displaystyle \cos 3\theta =4\cos ^{3}\theta -3\cos \theta \,} 半角公式 cos θ 2 = ± 1 + cos θ 2 . {\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}}.\,} 冪簡約公式 cos 2 θ = 1 + cos 2 θ 2 {\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}\,\!} cos 3 θ = 3 cos θ + cos 3 θ 4 {\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos 3\theta }{4}}\,\!} 和差化積公式 cos θ + cos ϕ = 2 cos ( θ + ϕ 2 ) cos ( θ − ϕ 2 ) {\displaystyle \cos \theta +\cos \phi =2\cos \left({\frac {\theta +\phi }{2}}\right)\cos \left({\frac {\theta -\phi }{2}}\right)} cos θ − cos ϕ = − 2 sin ( θ + ϕ 2 ) sin ( θ − ϕ 2 ) {\displaystyle \cos \theta -\cos \phi =-2\sin \left({\theta +\phi \over 2}\right)\sin \left({\theta -\phi \over 2}\right)} 萬能公式 cos α = 1 − tan 2 α 2 1 + tan 2 α 2 {\displaystyle \cos \alpha ={\frac {1-\tan ^{2}{\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}\,\!} ∫ cos c x d x = 1 c sin c x {\displaystyle \int \cos cx\;dx={\frac {1}{c}}\sin cx\,\!} ∫ cos n c x d x = cos n − 1 c x sin c x n c + n − 1 n ∫ cos n − 2 c x d x ( n > 0 ) {\displaystyle \int \cos ^{n}cx\;dx={\frac {\cos ^{n-1}cx\sin cx}{nc}}+{\frac {n-1}{n}}\int \cos ^{n-2}cx\;dx\qquad {\mbox{(}}n>0{\mbox{)}}\,\!} ∫ x cos c x d x = cos c x c 2 + x sin c x c {\displaystyle \int x\cos cx\;dx={\frac {\cos cx}{c^{2}}}+{\frac {x\sin cx}{c}}\,\!} ∫ x n cos c x d x = x n sin c x c − n c ∫ x n − 1 sin c x d x {\displaystyle \int x^{n}\cos cx\;dx={\frac {x^{n}\sin cx}{c}}-{\frac {n}{c}}\int x^{n-1}\sin cx\;dx\,\!} ∫ − a 2 a 2 x 2 cos 2 n π x a d x = a 3 ( n 2 π 2 − 6 ) 24 n 2 π 2 ( n = 1 , 3 , 5... ) {\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(}}n=1,3,5...{\mbox{)}}\,\!} ∫ cos c x x d x = ln | c x | + ∑ i = 1 ∞ ( − 1 ) i ( c x ) 2 i 2 i ⋅ ( 2 i ) ! {\displaystyle \int {\frac {\cos cx}{x}}dx=\ln |cx|+\sum _{i=1}^{\infty }(-1)^{i}{\frac {(cx)^{2i}}{2i\cdot (2i)!}}\,\!} ∫ cos c x x n d x = − cos c x ( n − 1 ) x n − 1 − c n − 1 ∫ sin c x x n − 1 d x ( n ≠ 1 ) {\displaystyle \int {\frac {\cos cx}{x^{n}}}dx=-{\frac {\cos cx}{(n-1)x^{n-1}}}-{\frac {c}{n-1}}\int {\frac {\sin cx}{x^{n-1}}}dx\qquad {\mbox{(}}n\neq 1{\mbox{)}}\,\!} ∫ d x cos c x = 1 c ln | tan ( c x 2 + π 4 ) | {\displaystyle \int {\frac {dx}{\cos cx}}={\frac {1}{c}}\ln \left|\tan \left({\frac {cx}{2}}+{\frac {\pi }{4}}\right)\right|} ∫ d x cos n c x = sin c x c ( n − 1 ) c o s n − 1 c x + n − 2 n − 1 ∫ d x cos n − 2 c x ( n > 1 ) {\displaystyle \int {\frac {dx}{\cos ^{n}cx}}={\frac {\sin cx}{c(n-1)cos^{n-1}cx}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}cx}}\qquad {\mbox{(}}n>1{\mbox{)}}\,\!} ∫ d x 1 + cos c x = 1 c tan c x 2 {\displaystyle \int {\frac {dx}{1+\cos cx}}={\frac {1}{c}}\tan {\frac {cx}{2}}\,\!} ∫ d x 1 − cos c x = − 1 c cot c x 2 {\displaystyle \int {\frac {dx}{1-\cos cx}}=-{\frac {1}{c}}\cot {\frac {cx}{2}}\,\!} ∫ x d x 1 + cos c x = x c tan c x 2 + 2 c 2 ln | cos c x 2 | {\displaystyle \int {\frac {x\;dx}{1+\cos cx}}={\frac {x}{c}}\tan {\frac {cx}{2}}+{\frac {2}{c^{2}}}\ln \left|\cos {\frac {cx}{2}}\right|} ∫ x d x 1 − cos c x = − x c cot c x 2 + 2 c 2 ln | sin c x 2 | {\displaystyle \int {\frac {x\;dx}{1-\cos cx}}=-{\frac {x}{c}}\cot {\frac {cx}{2}}+{\frac {2}{c^{2}}}\ln \left|\sin {\frac {cx}{2}}\right|} ∫ cos c x d x 1 + cos c x = x − 1 c tan c x 2 {\displaystyle \int {\frac {\cos cx\;dx}{1+\cos cx}}=x-{\frac {1}{c}}\tan {\frac {cx}{2}}\,\!} ∫ cos c x d x 1 − cos c x = − x − 1 c cot c x 2 {\displaystyle \int {\frac {\cos cx\;dx}{1-\cos cx}}=-x-{\frac {1}{c}}\cot {\frac {cx}{2}}\,\!} ∫ cos c 1 x cos c 2 x d x = sin ( c 1 − c 2 ) x 2 ( c 1 − c 2 ) + sin ( c 1 + c 2 ) x 2 ( c 1 + c 2 ) ( | c 1 | ≠ | c 2 | ) {\displaystyle \int \cos c_{1}x\cos c_{2}x\;dx={\frac {\sin(c_{1}-c_{2})x}{2(c_{1}-c_{2})}}+{\frac {\sin(c_{1}+c_{2})x}{2(c_{1}+c_{2})}}\qquad {\mbox{(}}|c_{1}|\neq |c_{2}|{\mbox{)}}\,\!} More information , ... 弳度 0 {\displaystyle 0} π 12 {\displaystyle {\frac {\pi }{12}}} π 10 {\displaystyle {\frac {\pi }{10}}} π 6 {\displaystyle {\frac {\pi }{6}}} π 5 {\displaystyle {\frac {\pi }{5}}} π 4 {\displaystyle {\frac {\pi }{4}}} 3 π 10 {\displaystyle {\frac {3\pi }{10}}} π 3 {\displaystyle {\frac {\pi }{3}}} 2 π 5 {\displaystyle {\frac {2\pi }{5}}} 5 π 12 {\displaystyle {\frac {5\pi }{12}}} π 2 {\displaystyle {\frac {\pi }{2}}} 角度 0 ∘ {\displaystyle 0^{\circ }} 15 ∘ {\displaystyle 15^{\circ }} 18 ∘ {\displaystyle 18^{\circ }} 30 ∘ {\displaystyle 30^{\circ }} 36 ∘ {\displaystyle 36^{\circ }} 45 ∘ {\displaystyle 45^{\circ }} 54 ∘ {\displaystyle 54^{\circ }} 60 ∘ {\displaystyle 60^{\circ }} 72 ∘ {\displaystyle 72^{\circ }} 75 ∘ {\displaystyle 75^{\circ }} 90 ∘ {\displaystyle 90^{\circ }} cos 1 {\displaystyle 1} 6 + 2 4 {\displaystyle {\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}} 2 ( 5 + 5 ) 4 {\displaystyle {\frac {\sqrt {2\left(5+{\sqrt {5}}\right)}}{4}}} 3 2 {\displaystyle {\frac {\sqrt {3}}{2}}} 5 + 1 4 {\displaystyle {\frac {{\sqrt {5}}+1}{4}}} 2 2 {\displaystyle {\frac {\sqrt {2}}{2}}} 2 ( 5 − 5 ) 4 {\displaystyle {\frac {\sqrt {2\left(5-{\sqrt {5}}\right)}}{4}}} 1 2 {\displaystyle {\frac {1}{2}}} 5 − 1 4 {\displaystyle {\frac {{\sqrt {5}}-1}{4}}} 6 − 2 4 {\displaystyle {\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}} 0 {\displaystyle 0} Close More information , ... 角度 0 ∘ {\displaystyle 0^{\circ }} 30 ∘ {\displaystyle 30^{\circ }} 45 ∘ {\displaystyle 45^{\circ }} 60 ∘ {\displaystyle 60^{\circ }} 90 ∘ {\displaystyle 90^{\circ }} cos 4 2 = 1 {\displaystyle {\frac {\sqrt {4}}{2}}=1} 3 2 {\displaystyle {\frac {\sqrt {3}}{2}}} 2 2 {\displaystyle {\frac {\sqrt {2}}{2}}} 1 2 = 1 2 {\displaystyle {\frac {\sqrt {1}}{2}}={1 \over 2}} 0 2 = 0 {\displaystyle {\frac {\sqrt {0}}{2}}=0} Close 主條目:餘弦定理 餘弦定理(也叫做餘弦公式)是勾股定理的擴展: c 2 = a 2 + b 2 − 2 a b cos C {\displaystyle c^{2}=a^{2}+b^{2}-2ab\cos C\,} 也表示為: cos C = a 2 + b 2 − c 2 2 a b {\displaystyle \cos C={\frac {a^{2}+b^{2}-c^{2}}{2ab}}\,\!} 這個定理也可以通過把三角形分為兩個直角三角形來證明。餘弦定律用於在一個三角形的兩個邊和一個角已知時確定未知的數據。 如果這個角不包含在這兩個邊之間,三角形可能不是唯一的(邊-邊-角全等歧義)。小心餘弦定律的這種歧義情況。 維基共享資源上的相關多媒體資源:餘弦 數學主題 正弦 正切 餘切 正割 餘割 三角學 三角函數 函數 正弦波 Wikiwand in your browser!Seamless Wikipedia browsing. 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