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abstract
- Discuss sin , cos , tan \sin,\cos,\tansin,cos,The image properties of tan , these properties can be analyzed with the help of the unit circle
- y = cos xy=\cos{x}y=cosx : cosine curve
- y = sin xy=\sin{x}y=sinx : sinusoidal curve
- y = tan xy=\tan{x}y=tanx :tangent curve
Sinusoids and sinusoids | |
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cosine curve | |
Tangent lines and tangent curves |
sine function
- y = sin xy=\sin{x}y=sinx, x ∈ R x\in\mathbb{R} x∈R is a sine function, where the independent variablexxx is the radian value
- Domain: R \mathbb{R}R
- Value range: [ − 1 , 1 ] [-1,1][−1,1]
- 当x = − π 2 + 2 k π x=-\frac{\pi}{2}+2k\pix=−2p+2kπ, k ∈ Z k\in\mathbb{Z} k∈Obtain the minimum value at Z − 1 -1−1
- 当x = π 2 + 2 k π x=\frac{\pi}{2}+2k\pix=2p+2kπ, k ∈ Z k\in\mathbb{Z} k∈Get the maximum value 1 when Z
- Boundedness: ∣ sin x ∣ ⩽ 1 |\sin{x}|\leqslant{1}∣sinx∣⩽1
- Parity: odd function
- Periodicity: The minimum positive period is 2 π 2\pi2 p.m
- Solution: [ − π 2 + 2 k π , π 2 + 2 k π ] [-\frac{\pi}{2}+2k\pi,\frac{\pi}{2}+2k\pi][−2p+2 kp ,2p+2 kπ ] equivalent; [ π 2 + 2 k π , 3 π 2 + 2 k π ] [\frac{\pi}{2}+2k\pi,\frac{\pi}{2}+2k\pi][2p+2 kp ,23 p.m+2 kπ ] is a decreasing interval;k ∈ Z k\in\mathbb{Z}k∈Z
- The sine function describes radians xxThe corresponding sine value of x is sin x \sin{x}sinx
sine function
- y = A sin ( ω x + ϕ ) y=A\sin(\omega{x}+\phi)y=Asin ( ω x+ϕ ) , called a sinusoidal function
- Compared to the sine function y = sin xy=\sin{x}y=sinx ,y = A sin ( ω x + ϕ ) y=A\sin(\omega{x}+\phi)y=Asin ( ω x+ϕ ) adds two parametersA , ω , ϕ A,\omega,\phi thatA,oh _ϕ (they are not variables, but constants)
- This function is very common in physical applications and has obvious physical meaning.
- The sum of sinusoidal functions can still be described by circular motion:
- Assume a certain point P (x, y) P(x,y) in the rectangular coordinate systemP(x,y ) around the originOOO takes radius asRRThe angular velocityof the circular trajectory of R isω \omegaCircular motion of ω rad/s
- Let PP before rotationThe position of P isP 0 P_0P0, 且OP OPOP isϕ \phiterminal edge of ϕ
- After ttAfter t seconds, clickPPP has arrived at a new location, andOP OPOP isϕ + ω t \phi+\omega{t}ϕ+Terminal edge of ω t
- Easy to calculate PPCoordinates of P (x, y) (x,y)(x,y ) about timettFunctional relationship of t
- x = R cos ( ω t + ϕ ) x=R\cos(\omega{t}+\phi)x=Rcos(ωt+) _
- y = R sin ( ω t + ϕ ) y=R\sin(\omega{t}+\phi)y=Rsin ( ω t+) _
- The derivation method is as follows:
- In the Cartesian coordinate system x O y xOyOn x O y , let the angleϕ \phiThe vertex of ϕ isOOThe origins of the O coordinates coincide with each other,ϕ \phiThe initial side of ϕ andxxPositive x- axis coincidence
- And with OOO is the center of the circle to construct the unit circle,ϕ \phiThe coordinates of the intersection of ϕ and the unit circle are E ( cos ϕ , sin ϕ ) E(\cos\phi,\sin{\phi})E ( cosϕ ,sin) _
- And P 0 P_0P0Also ϕ \phiϕThe point on the terminal edge, andOP 0 = R OP_0=ROP0=R ; thenP 0 P_0P0The coordinates of (P 0 x , P 0 y ) (P_{0x},P_{0y})(P0x _,P0 years) isEEThe coordinates of E ( sin α , cos α ) (\sin\alpha,\cos\alpha)(sina ,cosRRof α )Rfold :P 0 x = R cos ϕ P_{0x}=R\cos\phiP0x _=Rcosϕ ,P 0 y = R sin ϕ P_{0y}=R\sin\phiP0 years=Rsinϕ
- For the terminal side ω t + ϕ \omega{t}+\phiω t+PPon ϕThe coordinates of point P are( R cos ( ω t + ϕ ) , R sin ( ω t + ϕ ) ) (R\cos(\omega{t}+\phi),R\sin(\omega{t}+ \phi))(Rcos(ωt+ϕ ) ,Rsin ( ω t+ϕ ))
- This gives us the sinusoidal function y = A sin ( ω x + ϕ ) y=A\sin(\omega{x}+\phi)y=Asin ( ω x+) _
Rotation related concepts
Rotation angular velocity
- Point P ( x , y ) P(x,y) in the coordinate systemP(x,y ) pointOOThe number of radians of the angle that O rotates in unit timeω \omegaoh
rotation period
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y = R sin ( ω t + ϕ ) y=R\sin(\omega{t}+\phi)y=Rsin ( ω t+ϕ ) , pointPPThe time required for P to rotate once is T = 2 π ω T=\frac{2\pi}{\omega}T=oh2 p.m, this time is also called the rotation period
- 令α = ω t + ϕ \alpha=\omega{t}+\phia=ω t+ϕ , let the functionyyThe minimum positive period of y isT 0 T_0T0,则 y ( t + T 0 ) y(t+T_0) y(t+T0)= y ( t ) y(t) y(t)
- 即R sin ( ω ( t + T 0 ) + ϕ ) R\sin(\omega{(t+T_0)}+\phi)Rsin ( ω ( t+T0)+ϕ ) =R sin ( ω t + ϕ ) R\sin(\omega{t}+\phi)Rsin ( ω t+ϕ ) , 即sin ( ( ω t + ϕ ) + ω T 0 ) \sin((\omega{t}+\phi)+\omega T_0)sin (( ω t+) _+ω T0) =sin ( ω t + ϕ ) \sin(\omega{t}+\phi)sin ( ω t+) _
- So sin ( α + T 0 ) = sin ( α ) \sin(\alpha+T_0)=\sin(\alpha)sin ( a+T0)=sin ( a )
- sin α \sin\alphasinThe period of α is2 π 2\pi2 π , thenω T 0 \omega{T_0}ω T0= 2 π 2\pi2 π , soT 0 = 2 π ω T_{0}=\frac{2\pi}{\omega}T0=oh2 p.m
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In addition, the rotation period calculation formula can also be obtained from the expansion and contraction angle of the coordinates.
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Example: sin ( kx ) \sin(kx)sin(kx), k = 1 , 2 , 3 , ⋯ , n k=1,2,3,\cdots,n k=1,2,3,⋯,When n , at[ 0 , 2 π ] [0,2\pi][0,2 π ] to obtain the extreme point of the maximum value 1 (satisfyingkx = π 2 kx=\frac{\pi}{2}kx=2p) are: π 2 \frac{\pi}{2}2p, π 4 \frac{\pi}{4}4p, π 6 \frac{\pi}{6}6p, ⋯ \cdots ⋯ ,π 2 n \frac{\pi}{2n}2 np
Rotation frequency
- Within one second, click PPThe number of times P rotates f = 1 T f=\frac{1}{T}f=T1= ω 2 π \frac{\omega}{2\pi}2 p.moh, called the frequency of rotation
first appearance
- Angle ϕ \phiϕ is also calledthe first phase
summary
- The rotation period (rotation frequency) only sums up with ω \omegaω is related toϕ \phiϕirrelevant _
- ω \omegaThe larger ω is, the more times the curve fluctuates within a certain range, and vice versa.
Graph and properties of cosine function
- We can induce the formula y = sin ( π 2 + x ) y=\sin(\frac{\pi}{2}+x)y=sin(2p+x)= cos x \cos{x} cosx得知,y = cos xy=\cos{x}y=cosGraph of x andsin ( π 2 + x ) \sin(\frac{\pi}{2}+x)sin(2p+The image of x ) is the same
- Therefore, we can study the cosine function by studying the sine function (the image of the cosine function can be obtained by shifting the sine function 2 units to the left)
- In addition, the cosine function y = A cos ( ω x + ϕ ) y=A\cos(\omega{x}+\phi)y=Acos ( ω x+ϕ ) can be converted toy = A sin ( ω x + ϕ + π 2 ) y=A\sin(\omega{x}+\phi+\frac{\pi}{2})y=Asin ( ω x+ϕ+2p)
nature
- The domain, range and period are the same as the sine function
- If and only if x = π + 2 k π x=\pi+2k\pix=Pi+2kπ, k ∈ Z k\in\mathbb{Z} k∈When Z , obtain the minimum value− 1 -1−1
- If and only if x = 2 k π x=2k\pix=2kπ, k ∈ Z k\in\mathbb{Z} k∈When Z , the cosine function takes the maximum value 1
- Parity: even function
- Form: [ 2 k π , π + 2 k π ] [2k\pi,\pi+2k\pi][ 2 kp ,Pi+2kπ], k ∈ Z k\in\mathbb{Z} k∈The function at Z is monotonically decreasing; [ π + 2 k π , 2 π + 2 k π ] [\pi+2k\pi,2\pi+2k\pi][ p+2 kp ,2 p.m+2kπ], k ∈ Z k\in\mathbb{Z} k∈The Z function is monotonically increasing
Graph and properties of tangent function
- Form: { x ∣ x ≠ k π + π 2 , k ∈ Z } \set{x|x\neq{k\pi+\frac{\pi}{2}},k\in\mathbb{Z}}{ x∣x=kπ+2p,k∈Z},
- Range: R \mathbb{R}R
- In the interval ( − π 2 , π 2 ) (-\frac{\pi}{2},\frac{\pi}{2})(−2p,2p) , whenx < π 2 x<\frac{\pi}{2}x<2pAnd infinitely close to π 2 \frac{\pi}{2}2ptime, tan x \tan{x}tanx tends to infinity, recorded astan x → + ∞ \tan{x}\to{+\infin}tanx→+∞
- The other side has x → − π 2 x\to{-\frac{\pi}{2}}x→−2ptime x → − ∞ \tan{x}\to{-\infin}tanx→−∞
- Period: π \piPi
- Yutan ( π + x ) \tan(\pi+x)tan ( π+x ) =tan x \tan{x}tanx , soπ \piπ istan x \tan{x}tana period of x
- And combined with the sinusoidal line in the unit circle, it is easy to show that the minimum positive period is π \piPi
- Parity: odd function
- Solution: ( − π 2 + k π , π 2 + k π ) (-\frac{\pi}{2}+k\pi,\frac{\pi}{2}+k\pi)(−2p+kπ ,2p+kπ), k ∈ Z k\in\mathbb{Z} k∈The function in the Z interval increases monotonically
Find angles from known trigonometric function values
Within any angle range
- The unit circle can often be used to solve for angles in radians with given values of trigonometric functions.
- For example: It is known that sin x = 2 2 \sin{x}=\frac{\sqrt{2}}{2}sinx=22,beg
- x x Possible values of x
- x ∈ [ − π 2 , π 2 ] x\in[-\frac{\pi}{2},\frac{\pi}{2}]x∈[−2p,2p] Conditionxxvalue of x
- untie:
- It can be seen from the unit circle that π 4 + 2 k π \frac{\pi}{4}+2k\pi4p+2 kπ ,( π − π 4 ) + 2 k π (\pi-\frac{\pi}{4})+2k\pi( p−4p)+2kπ, ( k ∈ Z ) (k\in\mathbb{Z}) (k∈Z ) all satisfysin x = 2 2 \sin{x}=\frac{\sqrt{2}}{2}sinx=22,
- Expressed as a set: { x ∣ x = 2 k π + π 4 ( k ∈ Z ) } \set{x|x=2k\pi+\frac{\pi}{4}(k\in{\mathbb{Z }})}{ x∣x=2 kp+4p(k∈Z)} ⋃ \bigcup⋃ { x ∣ x = 2 k π + 3 π 4 ( k ∈ Z ) } \set{x|x=2k\pi+\frac{3\pi}{4}(k\in\mathbb{Z})} { x∣x=2 kp+43 p.m(k∈Z)}
- 若x ∈ [ − π 2 , π 2 ] x\in[-\frac{\pi}{2},\frac{\pi}{2}]x∈[−2p,2p] , from the unit circle,x = π 4 x=\frac{\pi}{4}x=4p
- It can be seen from the unit circle that π 4 + 2 k π \frac{\pi}{4}+2k\pi4p+2 kπ ,( π − π 4 ) + 2 k π (\pi-\frac{\pi}{4})+2k\pi( p−4p)+2kπ, ( k ∈ Z ) (k\in\mathbb{Z}) (k∈Z ) all satisfysin x = 2 2 \sin{x}=\frac{\sqrt{2}}{2}sinx=22,
Inverse trigonometric function (within limited range)
- The function established by the process of finding angles with known trigonometric function values is called inverse trigonometric function
- Since only injective functions have inverse functions, inverse trigonometric functions are defined according to a monotonic interval within the restricted trigonometric function.
arcsine
- Generally, for the sine function y = sin xy=\sin{x}y=sinx , if the known function value isy ( y ∈ [ − 1 , 1 ] ) y(y\in[-1,1])and ( and∈[−1,1 ]) , then[ − π 2 , π 2 ] [-\frac{\pi}{2},\frac{\pi}{2}][−2p,2p] by the one and onlyxxThe x value corresponds to it, recorded asx = arcsin yx=\arcsin{y}x=arcsiny ,(其中− 1 ⩽ y ⩽ 1 -1\leqslant{y}\leqslant{1}−1⩽y⩽1 ,− π 2 ⩽ x ⩽ π 2 -\frac{\pi}{2}\leqslant{x}\leqslant{\frac{\pi}{2}}−2p⩽x⩽2p)
- For example: sin x = 1 2 \sin{x}=\frac{1}{2}sinx=21, x ∈ [ − π 2 , π 2 ] x\in[-\frac{\pi}{2},\frac{\pi}{2}]x∈[−2p,2p] , pleasexxThe problem of x can be expressed asarcsin 1 2 \arcsin{\frac{1}{2}}arcsin21
Arc cosine
- In the interval [0, π] [0,\pi][0,π ] meets the conditioncos x = y \cos{x}=ycosx=y, ( y ∈ [ − 1 , 1 ] ) (y\in[-1,1]) (y∈[−1,1 ]) to find anglexxx , recorded asx = arccos yx=\arccos{y}x=arccosy
- example
- arccos 1 2 = π 3 \arccos{\frac{1}{2}}=\frac{\pi}{3}arccos21=3p
- It is known that cos x = − 2 2 \cos{x}=-\frac{\sqrt{2}}{2}cosx=−22,且x ∈ [ 0 , 2 π ] x\in[0,2\pi]x∈[0,2 π ] , findxxThe value set of x
- It can be seen from the unit circle that the solution set is { 3 π 4 , 5 π 4 } \set{\frac{3\pi}{4},\frac{5\pi}{4}}{ 43 p.m,45 p.m}
- Expressed by the inverse cosine function: { arccos ( − 2 2 ) , π + arccos 2 2 } \set{\arccos{(-\frac{\sqrt{2}}{2})},\pi+\arccos\ frac{\sqrt{2}}{2}}{ arccos(−22),Pi+arccos22}
arctangent
- For example, write tan x = y ( y ∈ R ) \tan{x}=y(y\in\mathbb{R})tanx=and ( and∈R ) , 且x ∈ ( − π 2 , π 2 ) x\in(-\frac{\pi}{2},\frac{\pi}{2})x∈(−2p,2p) , then for each tangent valueyyy , in the open interval( − π 2 , π 2 ) (-\frac{\pi}{2},\frac{\pi}{2})(−2p,2p) , there is and is only one anglexxx fulltan x = y \tan{x}=ytanx=y ,definitionx = arctan yx=\arctan{y}x=arctany ,x ∈ ( − π 2 , π 2 ) x\in(-\frac{\pi}{2},\frac{\pi}{2})x∈(−2p,2p)
- For example arctan 3 3 \arctan{\frac{\sqrt{3}}{3}}arctan33= π 6 \frac{\pi}{6}6p