EM@trigonometric function induction formula@trigonometric function formula simplification

induction formula

  • The trigonometric functions of acute angles are simple and easy to find (easy to express)
  • For any angle, there are certain relationships between its various trigonometric functions that need to be discussed
  • The most basic and commonly used formula for inducing trigonometric functions

Same end angles

  • In the Cartesian coordinate system, α , α + 2 k π \alpha,\alpha+2k\pia ,a+2, k ∈ K k\in\mathbb{K} kIf the terminal sides of K are the same, they are defined by trigonometric functions. It is easy to know that the two trigonometric functions are equal.
    • cos ⁡ ( α + 2 k π ) = cos ⁡ α \cos(\alpha+2k\pi)=\cos\alphacos ( a+2 kp )=cosa
    • sin ⁡ ( α + 2 k π ) = sin ⁡ α \sin(\alpha+2k\pi)=\sin\alphasin ( a+2 kp )=sina
    • tan ⁡ ( α + 2 k π ) = tan ⁡ α \tan(\alpha+2k\pi)=\tan\alphatan ( a+2 kp )=tana
Internalization of trigonometric functions for any angle
  • According to the relationship between the trigonometric functions of the same terminal angles, all absolute values ​​exceed one cycle ( 2 π 2\pi2 π or− 2 π -2\pi2 π ) can be converted into an absolute value less than2 π 2\pi2 π angle to calculate

Opposite angle

  • About xxAngle that is symmetrical about the x- axis

  • a \alphaThe opposite angle of α is− α -\alphaa

  • Obviously, the terminal sides of opposite angles are about xxThe x- axis is symmetrical. According to the definition of trigonometric functions, we have

    • cos ⁡ ( − α ) \cos(-\alpha)cos ( α ) =cos ⁡ α \cos\alphacosa

    • sin ⁡ ( − α ) \sin(-\alpha)sin ( α ) =− sin ⁡ α -\sin\alphasina

    • tan ⁡ ( − α ) \tan(-\alpha)tan ( α ) =sin ⁡ ( − α ) / cos ⁡ ( − α ) \sin(-\alpha)/\cos(-\alpha)sin ( α ) /cos ( α ) =− tan ⁡ α -\tan\alphatana

  • summary

    • cos ⁡ α \cos\alphacosα is an even function, andsin ⁡ α , tan ⁡ α \sin\alpha,\tan\alphasina ,tanα is an odd function
Negative angle normalization of any angle
  • From the conclusion of opposite angles, any negative angle can be converted into a positive angle to calculate and express
  • For example cos ⁡ ( − π 4 ) \cos(-\frac{\pi}{4})cos(4p) =cos ⁡ π 4 \cos\frac{\pi}{4}cos4p; sin ⁡ ( − 7 π 3 ) \sin(-\frac{7\pi}{3})sin(37 p.m) =− sin ⁡ 7 π 3 -\sin\frac{7\pi}{3}sin37 p.m, tan ⁡ ( − π 3 ) \tan(-\frac{\pi}{3})tan(3p) =− tan ⁡ π 3 -\tan\frac{\pi}{3}tan3p

Origin symmetry angle

  • Angle α \alphaThe starting side of α is xxIn the rectangular coordinate system of the positive x -axis,α \alphaThe angle corresponding to the terminal side of α with respect to the terminal side of the origin is expressed as α + ( 2 k + 1 ) π \alpha+(2k+1)\pia+( 2k _+1 ) πψα + ( 2 k − 1 ) π \alpha+(2k-1)\pia+( 2k _1)π, k ∈ Z k\in\mathbb{Z} kZ ; Let’s call this type of angleα \alphaThe origin symmetry angleof α

    • In [ 0 , 2 π ) [0,2\pi)[0,α \alphawithin 2 π )The terminal side of α is symmetrical about the origin and is expressed asα ± π \alpha\pm{\pi}a±Pi
    • Then according to the formula for generating angles with the same terminal side, we get the expression for the symmetrical angle at the origin.
    • The coordinate signs of the points on the two terminal edges that are symmetrical about the origin are inverted.
  • a \alphaThe symmetry angle between α and its originα + ( 2 k + 1 ) π \alpha+(2k+1)\pia+( 2k _+1 ) Trigonometric function relationship of π :

    • cos ⁡ ( α + ( 2 k + 1 ) π ) \cos(\alpha+(2k+1)\pi)cos ( a+( 2k _+1 ) π ) =− cos ⁡ α -\cos\alphacosa

    • sin ⁡ ( α + ( 2 k + 1 ) π ) \sin(\alpha+(2k+1)\pi)sin ( a+( 2k _+1 ) π ) =− sin ⁡ α -\sin\alphasina

    • tan ⁡ ( α + ( 2 k + 1 ) π ) \tan(\alpha+(2k+1)\pi)tan ( a+( 2k _+1 ) π ) =tan ⁡ α \tan\alphatana

Any angle sharpening
  • Let the set of odd numbers be N 1 = { 2 k ∣ k ∈ Z } N_1=\set{2k|k\in{\mathbb{Z}}}N1={ 2k _kZ} , the even number set isN 2 = { 2 k + 1 ∣ k ∈ Z } N_2=\set{2k+1|k\in\mathbb{Z}}N2={ 2k _+1kZ}

    • sin ⁡ ( α + k π ) \sin(\alpha+k\pi)sin ( a+) ={ − sin ⁡ α k ∈ N 1 sin ⁡ α k ∈ N 2 \begin{cases}-\sin\alpha&k\in{N_1}\\ \sin\alpha&k\in{N_2}\end{cases}{ sinasinakN1kN2

    • cos ⁡ ( α + k π ) \cos(\alpha+k\pi)cos ( a+) ={ − cos ⁡ α k ∈ N 1 cos ⁡ α k ∈ N 2 \begin{cases}-\cos\alpha&k\in{N_1}\\ \cos\alpha&k\in{N_2}\end{cases}{ cosacosakN1kN2

    • tan ⁡ ( α + k π ) \tan(\alpha+k\pi)tan ( a+) =tan ⁡ α \tan\alphatanα, k ∈ Z k\in\mathbb{Z} kZ

  • After the above analysis and discussion, it can be seen that any angle can be transformed into α + k π \alpha+k\pia+ ,( ∣ α ∣ ⩽ π 2 ) (|\alpha|\leqslant\frac{\pi}{2})(α2p) form

  • Then according to α , − α \alpha,-\alphaa ,− The trigonometric function relationship of α is further transformed into[ 0 , π 2 ] [0,\frac{\pi}{2}][0,2p] to represent and calculate acute angle trigonometric functions

summary

  • The first three sets of formulas above: (three terminal edge relationships correspond to three sets of formulas)
    1. Same end angles
    2. Opposite angle
    3. Origin symmetry angle
  • Collectively called induced formulas , the formulas can be reasoned and memorized with the help of the terminal edge of any angle.
  • The induction formula can be used to find the value of a trigonometric function or simplify a trigonometric function.

complementary angle

  • Two angles are complementary ( α , π − α ) (\alpha,\pi-\alpha)( a ,Piα ) , then their terminal sides are aboutyyy- axis symmetry

    • π − α \pi-\alphaPiα can be viewed as− α + π -\alpha+\pia+π , that is, first aboutxxDraw− α -\alpha symmetrically about the x- axisα final edge, and then make− α -\alphaα is symmetric about the origin− α + π -\alpha+\pia+Pi
    • Press α \alpha respectivelyWhen the terminal side of α is in four quadrants, it is proved that the same conclusion can be obtained: α , π − α \alpha,\pi-\alphaa ,PiαAbout yy_y- axis symmetry
  • Known α \alphaα ,π − α \pi-\alphaPiα are complementary angles, thensin ⁡ ( π − α ) \sin(\pi-\alpha)sin ( pα ) =sin ⁡ α \sin\alphasina ? cos ⁡ ( π − α ) \cos(\pi-\alpha)cos ( pα ) =− cos ⁡ α -\cos\alphacosa

    • whereπ − α \pi-\alphaPiα is equivalent toα \alphaαAbout xx_After symmetry on the x- axis, then symmetry about the origin
    • sin ⁡ ( π − α ) \sin(\pi-\alpha)sin ( pα ) =− sin ⁡ ( − α ) -\sin(-\alpha)sin ( α ) =− ( − sin ⁡ α ) -(-\sin\alpha)(sinα ) =sin ⁡ α \sin\alphasina
    • Similarly, cos ⁡ ( π − α ) \cos(\pi-\alpha)cos ( pα ) =− cos ⁡ ( − α ) -\cos(-\alpha)cos ( α ) =− cos ⁡ α -\cos\alphacosa
  • In short, the sine values ​​of two complementary angles are equal, and the cosine values ​​are opposite to each other.

π 2 \frac{\pi}{2}2pTrigonometric functions related to angle expressions

  • Formula (the latter two can convert α \alpha in the first twoα is replaced by− α -\alphaα is obtained,5 ∼ 8 5\sim{8}58 can be directly obtained from the ratio relationship between each and the first two formulas)
    1. cos ⁡ ( α + π 2 ) \cos(\alpha+\frac{\pi}{2})cos ( a+2p) =− sin ⁡ α -\sin\alphasina ?
    2. sin ⁡ ( α + π 2 ) \sin(\alpha+\frac{\pi}{2})sin ( a+2p) =cos ⁡ α \cos\alphacosa
    3. cos ⁡ ( − α + π 2 ) \cos(-\alpha+\frac{\pi}{2})cos ( a+2p) =sin ⁡ α \sin\alphasina ?
    4. sin ⁡ ( − α + π 2 ) \sin(-\alpha+\frac{\pi}{2})sin ( a+2p) =cos ⁡ α \cos\alphacosa
    5. tan ⁡ ( α + π 2 ) \tan(\alpha+\frac{\pi}{2})tan ( a+2p)= − cot ⁡ α -\cot{\alpha} cota
    6. cot ⁡ ( α + π 2 ) \cot(\alpha+\frac{\pi}{2})cot ( a+2p) =− tan ⁡ α -\tan\alphatana
    7. tan ⁡ ( − α + π 2 ) = cot ⁡ α \tan(-\alpha+\frac{\pi}{2})=\cot{\alpha}tan ( a+2p)=cota
    8. cot ⁡ ( − α + π 2 ) \cot(-\alpha+\frac{\pi}{2})cot ( a+2p) =tan ⁡ α \tan\alphatana

α , α + π 2 \alpha,\alpha+\frac{\pi}{2}a ,a+2p

  • Discuss α \alphaαα + π 2 \alpha+\frac{\pi}{2}a+2ptrigonometric function relationship, we use
    • The coordinates of the intersection point of the terminal side and the unit circle (the horizontal and vertical coordinates respectively reflect the angle α \alphasine and cosine of α )
    • And the straight line y = ± xy=\pm x on the Cartesian coordinate systemy=± xauxiliary (transition)
      • P ( x , y ) P(x,y) P(x,y )Q ( y , x ) Q(y,x)Q(y,x ) with respect toy = xy=xy=xsymmetry _
        • For example (2, 1) (2,1)(2,1), ( 1 , 2 ) (1,2) (1,2 ) ; also as( 1 , − 2 ) (1,-2)(1,2), ( − 2 , 1 ) (-2,1) (2,1)
      • P ( x , y ) P(x,y)P(x,y )Q ( − y , − x ) Q(-y,-x)Q(y,x ) abouty = − xy=-xy=−xsymmetry _ _
        • For example ( 2 , 1 ) , ( − 1 , − 2 ) (2,1),(-1,-2)(2,1),(1,2 ) ; also as( 1 , − 2 ) (1,-2)(1,2), ( 2 , − 1 ) (2,-1) (2,1)
    • Coordinate relationships of points symmetrical about coordinate axes
      • P ( x , y ) , Q ( x , − y ) P(x,y),Q(x,-y) P(x,y),Q(x,y ) aboutxxx- axis symmetry
      • P ( x , y ) , Q ( − x , y ) P(x,y),Q(-x,y) P(x,y),Q(x,y ) Aboutyyy- axis symmetry
    • In fact, α \alphaα can be transformed toα + π 2 \alpha+\frac{\pi}{2}through two appropriateaxial symmetry transformationsa+2p
  • Let α \alphaThe terminal side α and the unit circle intersect at pointP ( cos ⁡ α , sin ⁡ α ) P(\cos\alpha,\sin\alpha)P(cosa ,sina )
  • Take α \alphaDiscussion of the first quadrant angle of α formula as an example
    • The first axial symmetry transformation is about the straight line y = xy=xy=x , the new coordinates obtained are denoted asMMM , from symmetry we can knowM ( sin ⁡ α , cos ⁡ α ) M(\sin\alpha,\cos\alpha)M(sina ,cosα ) , terminal edgeOM OMThe corresponding angle of OM : ( π 2 − α ) + 2 k π , k ∈ Z (\frac{\pi}{2}-\alpha)+2k\pi,k\in\mathbb{Z}(2pa )+2 kp ,kZ
    • The second axial symmetry transformation is about x = 0 x=0x=0 , the new coordinates obtained areNNN , byNNN andMMM with respect tox = 0 x=0x=0 symmetry, soN ( − sin ⁡ α , cos ⁡ α ) N(-\sin\alpha,\cos\alpha)N(sina ,cosa ) ;角α + π 2 \alpha+\frac{\pi}{2}a+2pThe terminal edge is ON ONON ,N ( cos ⁡ ( α + π 2 ) , sin ⁡ ( α + π 2 ) ) N(\cos(\alpha+\frac{\pi}{2}),\sin(\alpha+\frac{\pi }{2}))N ( cos ( a+2p),sin ( a+2p))
    • So cos ⁡ ( α + π 2 ) \cos(\alpha+\frac{\pi}{2})cos ( a+2p) =− sin ⁡ α -\sin\alphasina ? sin ⁡ ( α + π 2 ) \sin(\alpha+\frac{\pi}{2})sin ( a+2p) =cos ⁡ α \cos\alphacosa
  • Using similar techniques, α \alpha can be completely summarizedThe same conclusion (formula) is established when α is in the four quadrants.

α , α − π 2 \alpha,\alpha-\frac{\pi}{2}a ,a2p

  • α − π 2 \alpha-\frac{\pi}{2}a2p= − ( − α + π 2 ) -(-\alpha+\frac{\pi}{2})( a+2p)
  • It can be directly deduced from the formulas in the previous group, for example
    • cos ⁡ ( α − π 2 ) \cos(\alpha-\frac{\pi}{2})cos ( a2p) =cos ⁡ ( − ( − α + π 2 ) ) \cos(-(-\alpha+\frac{\pi}{2}))cos ( ( a+2p)) =cos ⁡ ( − α + π 2 ) \cos(-\alpha+\frac{\pi}{2})cos ( a+2p) =sin ⁡ α \sin\alphasina
    • sin ⁡ ( α − π 2 ) \sin(\alpha-\frac{\pi}{2})sin ( a2p) =sin ⁡ ( − ( − α + π 2 ) ) \sin(-(-\alpha+\frac{\pi}{2}))sin ( ( a+2p)) =− sin ⁡ ( − α + π 2 ) -\sin(-\alpha+\frac{\pi}{2})sin ( a+2p) =− cos ⁡ α -\cos\alphacosa
    • ⋯ \cdots

Summary@mantra

  • Through research and induction on the induced formulas of trigonometric functions, people have summarized a set of formulas to quickly complete the conversion of the following form

    • U ( α + k ⋅ π 2 ) U(\alpha+k\cdot{\frac{\pi}{2}})U ( a+k2p), k ∈ Z k\in\mathbb{Z} kZ V ( α ) V(\alpha) V ( a )
    • Inside U , VU,VU,V representssin ⁡ , cos ⁡ \sin,\cossin,A function name in cos , U, VU,VU,V may take the same function name
  • Here is the most commonly used formula, mainly used for sin ⁡ , cos ⁡ \sin,\cossin,cos

  • " Odd changes to even, unchanged, symbols look at the quadrants "

    • Changes from odd to even remain unchanged:
      • ifkk _k is an even number, thenU, VU, VU,The function names of V are the same, for example, they are allsin ⁡ \sinsin or bothcos ⁡ \coscos , that is,the function name remains unchanged
      • ifkk _k is an odd number, the function name changes (sin ⁡ → cos ⁡ ; cos ⁡ → sin ⁡ \sin\to\cos;\cos\to{\sin}sincos;cossin)
    • Symbols look at the quadrants:
      • The sign refers to the plus or minus sign
      • Change α \alphaα is regarded as an acute angle, and then judgeα + k ⋅ π 2 \alpha+k\cdot{\frac{\pi}{2}}a+k2pThe quadrant where the terminal edge is located
      • The sign of the terminal side (corresponding angle) under the trigonometric function U is VVThe symbol of V is simplythe same sign
    • cos ⁡ ( − 19 π 4 ) \cos(-\frac{19\pi}{4})cos(419 p.m) =cos ⁡ ( 19 4 π ) \cos(\frac{19}{4}\pi)cos(419π ) =cos ⁡ ( 3 π 4 + 4 π ) \cos(\frac{3\pi}{4}+4\pi)cos(43 p.m+4 π ) =cos ⁡ 3 4 π \cos{\frac{3}{4}\pi}cos43π =cos ⁡ ( π 2 + π 4 ) \cos(\frac{\pi}{2}+\frac{\pi}{4})cos(2p+4p) =− sin ⁡ π 4 -\sin\frac{\pi}{4}sin4p= − 2 2 -\frac{\sqrt{2}}{2} 22
  • Other trigonometric functions can be converted into sin ⁡ , cos ⁡ \sin,\cossin,cos is calculated, so it is not necessary to remember

  • If necessary, you can refer to other information for other formulas.

refs

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Origin blog.csdn.net/xuchaoxin1375/article/details/133394722
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