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sin 2 x + cos 2 x = 1 \sin ^2 x + \cos ^2 x = 1 sin2x+cos2x=1
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tan x = sin x cos x \tan x = \dfrac{\sin x}{\cos x} tanx=cosxsinx
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cot x = 1 tan x = cos x sin x \cot x = \dfrac{1}{\tan x}=\dfrac{\cos x}{\sin x} cotx=tanx1=sinxcosx
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sec x = 1 cos x \sec x= \dfrac{1}{\cos x} secx=cosx1
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csc x = 1 sin x \csc x =\dfrac{1}{\sin x} cscx=sinx1
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tan 2 x = sec 2 − 1 = 1 cos 2 x − 1 = 1 − cos 2 x cos 2 x = sin 2 x cos 2 x \tan^2x=\sec^2-1=\dfrac{1}{\cos^2x}-1=\dfrac{1-\cos^2x}{\cos^2x}=\dfrac{\sin^2x}{\cos^2x} tan2x=sec2−1=cos2x1−1=cos2x1−cos2x=cos2xsin2x
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cot 2 = csc 2 x − 1 = 1 sin 2 x − 1 = 1 − sin 2 x sin 2 x = cos 2 x sin 2 x \cot^2=\csc^2x-1=\dfrac{1}{\sin^2x}-1=\dfrac{1-\sin^2x}{\sin^2x}=\dfrac{\cos^2x}{\sin^2x} cot2=csc2x−1=sin2x1−1=sin2x1−sin2x=sin2xcos2x
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cos x = sin ( x + π 2 ) \cos x=\sin(x+\dfrac{\pi}{2}) cosx=sin(x+2π)， sin x \sin x sinx 向左平移 π 2 \dfrac{\pi}{2} 2π. (左加右减)
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sin x = cos ( x − π 2 ) \sin x=\cos(x-\dfrac{\pi}{2}) sinx=cos(x−2π)
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cos x = cos ( − x ) \cos x= \cos(-x) cosx=cos(−x)，偶函数
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sin x = − sin ( − x ) \sin x = - \sin(-x) sinx=−sin(−x)，奇函数
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sin x = − sin ( x ± π ) \sin x= -\sin(x\pm\pi) sinx=−sin(x±π)， sin x \sin x sinx无论是向左、还是向右平移 π \pi π 个单位后，乘以-1，关于x轴对称之后函数图像不变.
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cos x = − cos ( x ± π ) \cos x = -\cos(x\pm\pi) cosx=−cos(x±π)
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arcsin x + arccos x = π 2 \arcsin x+\arccos x=\dfrac{\pi}{2} arcsinx+arccosx=2π.
倍(半)角公式
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cos ( A ± B ) = cos A ⋅ cos B ∓ sin A ⋅ sin B \cos(A\pm B)=\cos A\cdot\cos B \mp \sin A\cdot\sin B cos(A±B)=cosA⋅cosB∓sinA⋅sinB.
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sin ( A ± B ) = sin A ⋅ cos B ± cos A ⋅ sin B \sin(A\pm B)=\sin A\cdot\cos B \pm \cos A\cdot\sin B sin(A±B)=sinA⋅cosB±cosA⋅sinB.
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cos ( 2 A ) = cos 2 A − sin 2 A = 1 − 2 sin 2 A = 2 cos 2 A − 1 \cos(2A)=\cos^2A-\sin^2A=1-2\sin^2A=2\cos^2A-1 cos(2A)=cos2A−sin2A=1−2sin2A=2cos2A−1.
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cos A = cos 2 A 2 − sin 2 A 2 = 1 − 2 sin 2 A 2 = 2 cos 2 A 2 − 1 \cos A = \cos^2\dfrac{A}{2}-\sin^2\dfrac{A}{2}=1-2\sin^2\dfrac{A}{2}=2\cos^2\dfrac{A}{2}-1 cosA=cos22A−sin22A=1−2sin22A=2cos22A−1.
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sin ( 2 A ) = 2 sin A ⋅ cos A \sin(2A)=2\sin A\cdot\cos A sin(2A)=2sinA⋅cosA.
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sin A = 2 sin A 2 ⋅ cos A 2 \sin A = 2\sin\dfrac{A}{2}\cdot\cos\dfrac{A}{2} sinA=2sin2A⋅cos2A.
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tan 2 α = 2 tan α 1 − tan 2 α \tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha} tan2α=1−tan2α2tanα.
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tan α = 2 tan α 2 1 − tan 2 α 2 \tan\alpha=\dfrac{2\tan\dfrac{\alpha}{2}}{1-\tan^2\dfrac{\alpha}{2}} tanα=1−tan22α2tan2α.
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tan α 2 = 1 − cos α sin α = sin α 1 + cos α \tan\dfrac{\alpha}{2}=\dfrac{1-\cos\alpha}{\sin\alpha}=\dfrac{\sin\alpha}{1+\cos\alpha} tan2α=sinα1−cosα=1+cosαsinα.
tan 2 α = sin 2 α cos 2 α = 2 sin α ⋅ cos α 1 − 2 sin 2 α = 2 tan α 1 cos 2 α − 2 tan 2 α = 2 tan α 1 − sin 2 α cos 2 α − tan 2 α = 2 tan α 1 − tan 2 α . \tan2\alpha=\dfrac{\sin2\alpha}{\cos2\alpha}=\dfrac{2\sin\alpha\cdot\cos\alpha}{1-2\sin^2\alpha}=\dfrac{2\tan\alpha}{\dfrac{1}{\cos^2\alpha}-2\tan^2\alpha}=\dfrac{2\tan\alpha}{\dfrac{1-\sin^2\alpha}{\cos^2\alpha}-\tan^2\alpha}=\dfrac{2\tan\alpha}{1-\tan^2\alpha}. tan2α=cos2αsin2α=1−2sin2α2sinα⋅cosα=cos2α1−2tan2α2tanα=cos2α1−sin2α−tan2α2tanα=1−tan2α2tanα.
tan α = 2 tan α 2 1 − tan 2 α 2 \tan\alpha=\dfrac{2\tan\dfrac{\alpha}{2}}{1-\tan^2\dfrac{\alpha}{2}} tanα=1−tan22α2tan2α
tan α − tan α tan 2 α 2 = 2 tan α 2 \tan\alpha-\tan\alpha\tan^2\dfrac{\alpha}{2}=2\tan\dfrac{\alpha}{2} tanα−tanαtan22α=2tan2α
tan α ⋅ tan 2 α 2 + 2 tan α 2 − tan α = 0 \tan\alpha\cdot\tan^2\dfrac{\alpha}{2}+2\tan\dfrac{\alpha}{2}-\tan\alpha=0 tanα⋅tan22α+2tan2α−tanα=0
求根公式：
tan α 2 = − 2 ± 4 + 4 tan 2 α 2 tan α = − 1 ± sec α tan α = − cos α ± 1 sin α \tan\dfrac{\alpha}{2}=\dfrac{-2\pm\sqrt{4+4\tan^2\alpha}}{2\tan\alpha}=\dfrac{-1\pm\sec\alpha}{\tan\alpha}=\dfrac{-\cos\alpha\pm1}{\sin\alpha} tan2α=2tanα−2±4+4tan2α=tanα−1±secα=sinα−cosα±1
当 α ∈ ( 0 , π ) \alpha\in(0,\pi) α∈(0,π) 时， tan α 2 > 0 \tan\dfrac{\alpha}{2}>0 tan2α>0，而 − cos α + 1 sin α < 0 -\dfrac{\cos\alpha+1}{\sin\alpha}<0 −sinαcosα+1<0.
∴ tan α 2 = − cos α + 1 sin α \therefore \tan\dfrac{\alpha}{2}=-\dfrac{\cos\alpha+1}{\sin\alpha} ∴tan2α=−sinαcosα+1 不成立.
∴ tan α 2 = 1 − cos α sin α = ( 1 − cos α ) ⋅ ( 1 + cos α ) sin α + sin α ⋅ cos α = 1 − cos 2 α sin α + sin α ⋅ cos α = sin α 1 + cos α \therefore \tan\dfrac{\alpha}{2}=\dfrac{1-\cos\alpha}{\sin\alpha}=\dfrac{(1-\cos\alpha)\cdot(1+\cos\alpha)}{\sin\alpha+\sin\alpha\cdot\cos\alpha}=\dfrac{1-\cos^2\alpha}{\sin\alpha+\sin\alpha\cdot\cos\alpha}=\dfrac{\sin\alpha}{1+\cos\alpha} ∴tan2α=sinα1−cosα=sinα+sinα⋅cosα(1−cosα)⋅(1+cosα)=sinα+sinα⋅cosα1−cos2α=1+cosαsinα
正、余弦化切弦
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sin 2 x = 2 sin x ⋅ cos x = 2 tan x sec 2 x = 2 tan x 1 + tan 2 x \sin2x=2\sin x\cdot\cos x=\dfrac{2\tan x}{\sec^2x}=\dfrac{2\tan x}{1+\tan^2x} sin2x=2sinx⋅cosx=sec2x2tanx=1+tan2x2tanx.
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cos 2 x = cos 2 x − sin 2 x = 1 − tan 2 x sec 2 x = 1 − tan 2 x 1 + tan 2 x \cos2x=\cos^2x-\sin^2x=\dfrac{1-\tan^2x}{\sec^2x}=\dfrac{1-\tan^2x}{1+\tan^2x} cos2x=cos2x−sin2x=sec2x1−tan2x=1+tan2x1−tan2x.
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sin x = 2 tan x 2 1 + tan 2 x 2 \sin x=\dfrac{2\tan\dfrac{x}{2}}{1+\tan^2\dfrac{x}{2}} sinx=1+tan22x2tan2x.
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cos x = 1 − tan 2 x 2 1 + tan 2 x 2 \cos x=\dfrac{1-\tan^2\dfrac{x}{2}}{1+\tan^2\dfrac{x}{2}} cosx=1+tan22x1−tan22x.
辅助角公式
- sin α ⋅ a a 2 + b 2 − cos α ⋅ b a 2 + b 2 = sin α ⋅ cos β − cos α ⋅ sin β = sin ( α − β ) \sin\alpha\cdot\dfrac{a}{\sqrt{a^2+b^2}}-\cos\alpha\cdot\dfrac{b}{\sqrt{a^2+b^2}}=\sin\alpha\cdot\cos\beta-\cos\alpha\cdot\sin\beta=\sin(\alpha-\beta) sinα⋅a2+b2a−cosα⋅a2+b2b=sinα⋅cosβ−cosα⋅sinβ=sin(α−β).
其中，令 cos β = a a 2 + b 2 \cos\beta=\dfrac{a}{\sqrt{a^2+b^2}} cosβ=a2+b2a， sin β = b a 2 + b 2 \sin\beta=\dfrac{b}{\sqrt{a^2+b^2}} sinβ=a2+b2b，
则有 cos 2 β + sin 2 β = ( a a 2 + b 2 ) 2 + ( b a 2 + b 2 ) 2 = 1 \cos^2\beta+\sin^2\beta=\Big(\dfrac{a}{\sqrt{a^2+b^2}}\Big)^2+\Big(\dfrac{b}{\sqrt{a^2+b^2}}\Big)^2=1 cos2β+sin2β=(a2+b2a)2+(a2+b2b)2=1.
E m L 2 ω 2 + R 2 ⋅ ( R ⋅ sin ω t − L ω ⋅ cos ω t ) \dfrac{E_m}{L^2\omega^2+R^2}\cdot\big(R\cdot\sin\omega t-L\omega\cdot\cos\omega t\big) L2ω2+R2Em⋅(R⋅sinωt−Lω⋅cosωt)
= E m L 2 ω 2 + R 2 ⋅ ( sin ω t ⋅ R L 2 ω 2 + R 2 − cos ω t ⋅ L ω L 2 ω 2 + R 2 ) =\dfrac{E_m}{\sqrt{L^2\omega^2+R^2}}\cdot\big(\sin\omega t\cdot\dfrac{R}{\sqrt{L^2\omega^2+R^2}}-\cos\omega t\cdot\dfrac{L\omega}{\sqrt{L^2\omega^2+R^2}}\big) =L2ω2+R2Em⋅(sinωt⋅L2ω2+R2R−cosωt⋅L2ω2+R2Lω)
= E m L 2 ω 2 + R 2 ⋅ sin ( ω t − φ ) =\dfrac{E_m}{\sqrt{L^2\omega^2+R^2}}\cdot\sin(\omega t-\varphi) =L2ω2+R2Em⋅sin(ωt−φ).
其中 cos φ = R L 2 ω 2 + R 2 \cos\varphi=\dfrac{R}{\sqrt{L^2\omega^2+R^2}} cosφ=L2ω2+R2R， sin φ = L ω L 2 ω 2 + R 2 \sin\varphi=\dfrac{L\omega}{\sqrt{L^2\omega^2+R^2}} sinφ=L2ω2+R2Lω.
积化和差、和差化积
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