EM@Graphic properties of commonly used trigonometric functions (middle school part)

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
Insert image description here Sinusoids and sinusoids
Insert image description here cosine curve
Insert image description here Tangent lines and tangent curves

sine function

  • y = sin ⁡ xy=\sin{x}y=sinx, x ∈ R x\in\mathbb{R} xR 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]
      • x = − π 2 + 2 k π x=-\frac{\pi}{2}+2k\pix=2p+2, k ∈ Z k\in\mathbb{Z} kObtain the minimum value at Z − 1 -11
      • x = π 2 + 2 k π x=\frac{\pi}{2}+2k\pix=2p+2, k ∈ Z k\in\mathbb{Z} kGet the maximum value 1 when Z
    • Boundedness: ∣ sin ⁡ x ∣ ⩽ 1 |\sin{x}|\leqslant{1}sinx1
    • 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 ] 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 ] is a decreasing interval;k ∈ Z k\in\mathbb{Z}kZ
  • 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
  • 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

    • 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)
    • 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
  • In addition, the rotation period calculation formula can also be obtained from the expansion and contraction angle of the coordinates.

  • 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+2, k ∈ Z k\in\mathbb{Z} kWhen Z , obtain the minimum value− 1 -11
    • If and only if x = 2 k π x=2k\pix=2, k ∈ Z k\in\mathbb{Z} kWhen 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+2], k ∈ Z k\in\mathbb{Z} kThe 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+2], k ∈ Z k\in\mathbb{Z} kThe 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}}{ xx=+2p,kZ},
  • 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}}x2ptime ⁡ 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+,2p+), k ∈ Z k\in\mathbb{Z} kThe 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 ,( π − π 4 ) + 2 k π (\pi-\frac{\pi}{4})+2k\pi( p4p)+2, ( k ∈ Z ) (k\in\mathbb{Z}) (kZ ) 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 }})}{ xx=2 kp+4p(kZ)} ⋃ \bigcup {   x ∣ x = 2 k π + 3 π 4 ( k ∈ Z )   } \set{x|x=2k\pi+\frac{3\pi}{4}(k\in\mathbb{Z})} { xx=2 kp+43 p.m(kZ)}
    • 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

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}1y1 ,− π 2 ⩽ x ⩽ π 2 -\frac{\pi}{2}\leqslant{x}\leqslant{\frac{\pi}{2}}2px2p)
  • 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
    1. arccos ⁡ 1 2 = π 3 \arccos{\frac{1}{2}}=\frac{\pi}{3}arccos21=3p
    2. 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 ( andR ) , 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

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Origin blog.csdn.net/xuchaoxin1375/article/details/133502399
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