Trigonometry - Formulas, Identities, Functions and Problems
三角学:公式,恒等式,函数和难题
三角学是数学的一个重要分支,主要涉及角度的特定函数及其应用和计算。在数学中,总共有六种不同类型的三角函数: 正弦( sin \sin sin),余弦( cos \cos cos),正割( sec \sec sec),余割( cosec \cosec cosec),正切( tan \tan tan)和余切(cot)。这六种不同类型的三角函数象征着直角三角形的不同边的比率之间的关系。这些三角函数也可以被称为弧度函数,因为它们的值可以被描述为半径为 1 1 1 的圆的 x x x 和 y y y 坐标的比率,与标准位置的角保持联系。
sin ( θ ) = 对 边 斜 边 = O p p o s i t e H y p o t e n u s e cos ( θ ) = 邻 边 斜 边 = A d j a c e n t H y p o t e n u s e tan ( θ ) = 对 边 邻 边 = O p p o s i t e A d j a c e n t csc ( θ ) = 1 sin ( θ ) = H y p o t e n u s e O p p o s i t e sec ( θ ) = 1 cos ( θ ) = H y p o t e n u s e A d j a c e n t cot ( θ ) = 1 tan ( θ ) = A d j a c e n t O p p o s i t e \begin{array}{lll} \sin (\theta) = \frac{对边}{斜边}=\frac {Opposite}{Hypotenuse}\qquad & \cos (\theta) =\frac{邻边}{斜边}= \frac {Adjacent}{Hypotenuse} \qquad & \tan (\theta) =\frac{对边}{邻边}= \frac {Opposite}{Adjacent} \\ \csc (\theta) = \frac{1}{\sin(\theta)} = \frac {Hypotenuse}{Opposite} \qquad & \sec (\theta) = \frac{1}{\cos(\theta)} = \frac {Hypotenuse}{Adjacent} \qquad & \cot (\theta) = \frac{1}{\tan(\theta)} = \frac {Adjacent}{Opposite} \end{array} sin(θ)=斜边对边=HypotenuseOppositecsc(θ)=sin(θ)1=OppositeHypotenusecos(θ)=斜边邻边=HypotenuseAdjacentsec(θ)=cos(θ)1=AdjacentHypotenusetan(θ)=邻边对边=AdjacentOppositecot(θ)=tan(θ)1=OppositeAdjacent
三角函数对研究三角形、光、声或波非常重要,它们在不同领域和范围内的数值可以从下表得到。
表1: 三角函数在不同领域和范围内的值。
三角函数
定义域Domain
值域Range
sin x \sin x sinx
R \mathbb{R} R,周期函数
− 1 ⩽ sin x ⩽ 1 -1\leqslant \sin x\leqslant 1 −1⩽sinx⩽1
cos x \cos x cosx
R \mathbb{R} R,周期函数
− 1 ⩽ cos x ⩽ 1 -1\leqslant \cos x\leqslant 1 −1⩽cosx⩽1
tan x \tan x tanx
R ∖ { 2 n + 1 2 π , n ∈ Z } \mathbb{R}\setminus\{\dfrac{2n+1}{2}\pi,\; n\in \mathbb{Z}\} R∖{
22n+1π,n∈Z}
R \mathbb{R} R
csc x \csc x cscx
R ∖ { n π , n ∈ Z } \mathbb{R}\setminus\{n \pi, n\in \mathbb{Z}\} R∖{
nπ,n∈Z}
R ∖ { x : − 1 < x < 1 } \mathbb{R}\setminus \{x: -1\lt x\lt 1\} R∖{
x:−1<x<1}
sec x \sec x secx
R ∖ { 2 n + 1 2 π , n ∈ Z } \mathbb{R}\setminus\{\dfrac{2n+1}{2}\pi, n\in \mathbb{Z}\} R∖{
22n+1π,n∈Z}
R ∖ { x : − 1 < x < 1 } \mathbb{R}\setminus \{x:-1\lt x \lt 1\} R∖{
x:−1<x<1}
cot x \cot x cotx
R ∖ { n π , n ∈ Z } \mathbb{R}\setminus \{n \pi, n\in \mathbb{Z}\} R∖{
nπ,n∈Z}
sin 2 x = 2 sin x cos x \sin 2x = 2 \sin x \cos x sin2x=2sinxcosx
cos 2 x = cos 2 x – sin 2 x = 1 – 2 sin 2 x = 2 cos 2 x – 1 \cos 2x = \cos^{2}x – \sin^{2}x = 1 – 2 \sin^{2}x = 2 \cos^{2}x – 1 cos2x=cos2x–sin2x=1–2sin2x=2cos2x–1
tan 2 x = 2 tan x 1 − tan 2 x \tan 2x = \dfrac {2 \tan x}{1-\tan^{2}x} tan2x=1−tan2x2tanx
sin 3 x = 3 sin x – 4 sin 3 x \sin 3x = 3 \sin x – 4 \sin^{3}x sin3x=3sinx–4sin3x
cos 3 x = 4 cos 3 x – 3 cos x \cos 3x = 4 \cos^{3}x – 3 \cos x cos3x=4cos3x–3cosx
tan 3 x = 3 tan x – tan 3 x 1 − 3 tan 2 x \tan 3x = \dfrac {3 \tan x – \tan^{3}x}{1- 3\tan^{2}x} tan3x=1−3tan2x3tanx–tan3x
和差公式(不同角)
sin ( α + β ) = sin ( α ) cos ( β ) + cos ( α ) sin ( β ) \sin (\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
sin ( α − β ) = sin ( α ) cos ( β ) – cos ( α ) sin ( β ) \sin (\alpha - \beta) = \sin(\alpha) \cos(\beta) – \cos(\alpha) \sin(\beta) sin(α−β)=sin(α)cos(β)–cos(α)sin(β)
cos ( α + β ) = cos ( α ) cos ( β ) – sin ( α ) sin ( β ) \cos (\alpha + \beta) = \cos(\alpha) \cos(\beta) – \sin(\alpha) \sin(\beta) cos(α+β)=cos(α)cos(β)–sin(α)sin(β)
cos ( α – β ) = cos ( α ) cos ( β ) + sin ( α ) sin ( β ) \cos (\alpha – \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta) cos(α–β)=cos(α)cos(β)+sin(α)sin(β)
tan ( α + β ) = tan ( α ) + tan ( β ) 1 – tan ( α ) tan ( β ) \tan (\alpha + \beta) = \dfrac {\tan(\alpha)+\tan(\beta)}{1–\tan(\alpha) \tan(\beta)} tan(α+β)=1–tan(α)tan(β)tan(α)+tan(β)
tan ( α – β ) = tan ( α ) – tan ( β ) 1 + tan ( α ) tan ( β ) \tan (\alpha – \beta) = \dfrac{\tan(\alpha)–\tan(\beta)}{ 1+\tan(\alpha)\tan(\beta)} tan(α–β)=1+tan(α)tan(β)tan(α)–tan(β)
sin ( A + B + C ) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C . \sin (A+B+C) = \sin A\cos B\cos C + \cos A\sin B\cos C + \cos A \cos B \sin C – \sin A \sin B \sin C. sin(A+B+C)=sinAcosBcosC+cosAsinBcosC+cosAcosBsinC–sinAsinBsinC.
sin ( A + B + C ) = ( − sin ( B ) sin ( C ) + cos ( B ) cos ( C ) ) sin ( A ) + ( sin ( B ) cos ( C ) + sin ( C ) cos ( B ) ) cos ( A ) \sin (A+B+C) = \left(- \sin{\left(B \right)} \sin{\left(C \right)} + \cos{\left(B \right)} \cos{\left(C \right)}\right) \sin{\left(A \right)} + \left(\sin{\left(B \right)} \cos{\left(C \right)} + \sin{\left(C \right)} \cos{\left(B \right)}\right) \cos{\left(A \right)} sin(A+B+C)=(−sin(B)sin(C)+cos(B)cos(C))sin(A)+(sin(B)cos(C)+sin(C)cos(B))cos(A)
cos ( A + B + C ) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C . \cos (A+B+C) = \cos A \cos B \cos C – \cos A \sin B \sin C – \sin A \cos B \sin C – \sin A \sin B \cos C. cos(A+B+C)=cosAcosBcosC–cosAsinBsinC–sinAcosBsinC–sinAsinBcosC.
cos ( A + B + C ) = ( − sin ( B ) sin ( C ) + cos ( B ) cos ( C ) ) cos ( A ) − ( sin ( B ) cos ( C ) + sin ( C ) cos ( B ) ) sin ( A ) \cos (A+B+C) = \left(- \sin{\left(B \right)} \sin{\left(C \right)} + \cos{\left(B \right)} \cos{\left(C \right)}\right) \cos{\left(A \right)} - \left(\sin{\left(B \right)} \cos{\left(C \right)} + \sin{\left(C \right)} \cos{\left(B \right)}\right) \sin{\left(A \right)} cos(A+B+C)=(−sin(B)sin(C)+cos(B)cos(C))cos(A)−(sin(B)cos(C)+sin(C)cos(B))sin(A)
tan ( A + B + C ) = tan A + tan B + tan C – tan A tan B tan C 1 – tan A tan B – tan B tan C – tan A tan C \tan (A+B+C) =\dfrac { \tan A + \tan B + \tan C – \tan A \tan B \tan C}{ 1 – \tan A \tan B – \tan B \tan C – \tan A \tan C} tan(A+B+C)=1–tanAtanB–tanBtanC–tanAtanCtanA+tanB+tanC–tanAtanBtanC
tan ( A + B + C ) = − tan ( A ) tan ( B ) tan ( C ) + tan ( A ) + tan ( B ) + tan ( C ) − tan ( A ) tan ( B ) − tan ( A ) tan ( C ) − tan ( B ) tan ( C ) + 1 \tan (A+B+C) =\frac{- \tan{\left(A \right)} \tan{\left(B \right)} \tan{\left(C \right)} + \tan{\left(A \right)} + \tan{\left(B \right)} + \tan{\left(C \right)}}{- \tan{\left(A \right)} \tan{\left(B \right)} - \tan{\left(A \right)} \tan{\left(C \right)} - \tan{\left(B \right)} \tan{\left(C \right)} + 1} tan(A+B+C)=−tan(A)tan(B)−tan(A)tan(C)−tan(B)tan(C)+1−tan(A)tan(B)tan(C)+tan(A)+tan(B)+tan(C)
cot ( A + B + C ) = cot A cot B cot C – cot A – cot B – cot C cot A cot B + cot B cot C + cot A cot C – 1 \cot (A+B+C) = \dfrac {\cot A \cot B \cot C – \cot A–\cot B–\cot C}{\cot A \cot B + \cot B \cot C + \cot A \cot C – 1} cot(A+B+C)=cotAcotB+cotBcotC+cotAcotC–1cotAcotBcotC–cotA–cotB–cotC
cot ( A + B + C ) = cot ( A ) cot ( B ) cot ( C ) − cot ( A ) − cot ( B ) − cot ( C ) cot ( A ) cot ( B ) + cot ( A ) cot ( C ) + cot ( B ) cot ( C ) − 1 \cot (A+B+C) = \frac{\cot{\left(A \right)} \cot{\left(B \right)} \cot{\left(C \right)} - \cot{\left(A \right)} - \cot{\left(B \right)} - \cot{\left(C \right)}}{\cot{\left(A \right)} \cot{\left(B \right)} + \cot{\left(A \right)} \cot{\left(C \right)} + \cot{\left(B \right)} \cot{\left(C \right)} - 1} cot(A+B+C)=cot(A)cot(B)+cot(A)cot(C)+cot(B)cot(C)−1cot(A)cot(B)cot(C)−cot(A)−cot(B)−cot(C)
sin A = 1 csc A \sin A = \dfrac {1}{\csc A} sinA=cscA1
cos A = 1 sec A \cos A = \dfrac {1}{\sec A} cosA=secA1
sec A = 1 cos A \sec A = \dfrac{1}{\cos A} secA=cosA1
csc A = 1 sin A \csc A = \dfrac{1}{\sin A} cscA=sinA1
tan A = 1 cot A = sin A cos A \tan A = \dfrac {1}{\cot A} = \dfrac {\sin A}{\cos A } tanA=cotA1=cosAsinA
cot A = 1 tan A = cos A sin A \cot A = \dfrac {1}{\tan A} = \dfrac {\cos A}{\sin A } cotA=tanA1=sinAcosA
三角函数的最小正周期
sin ( x + 2 π ) = sin x \sin (x + 2\pi ) = \sin x sin(x+2π)=sinx
cos ( x + 2 π ) = cos x \cos (x + 2\pi ) = \cos x cos(x+2π)=cosx
tan ( x + π ) = tan x \tan (x + \pi ) = \tan x tan(x+π)=tanx
cot ( x + π ) = cot x \cot (x + \pi ) = \cot x cot(x+π)=cotx
三角函数的半角公式
sin x 2 = ± 1 − cos x 2 \sin \dfrac{x}{2} = ±\sqrt{\dfrac{1-\cos x}{2}} sin2x=±21−cosx
cos x 2 = ± 1 + cos x 2 \cos \dfrac{x}{2}= ±\sqrt{\dfrac{1+\cos x}{2}} cos2x=±21+cosx
tan x 2 = 1 − cos x 1 + cos x = 1 − cos x sin x = sin x 1 + cos x \tan \dfrac{x}{2}= \sqrt{\dfrac{1- \cos x}{1+ \cos x}} = \dfrac{1- \cos x}{\sin x} = \dfrac{\sin x}{1+\cos x} tan2x=1+cosx1−cosx=sinx1−cosx=1+cosxsinx
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