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}$ length is $l$, the vector and the $x$ axis and the $y$ axis are $\alpha$ and $\beta$ respectively, then we can easily get the coordinates of the vector is expressed as $\MATHBF = (l\cos\ Alpha, L\cos\beta) ^\text{t}$; Similarly, for three-dimensional space vector $\mathbf{b}$, its length is $l$, and the $x$ axis, the $y $ axis and the $z$ axis are $\alpha$, $\beta$ and $\gamma$, respectively, The coordin

Basic Relationship Between trigonometric functions of the same angle Reciprocal Relationship: Business Relationship: Square relationship: Tan α-cot α = 1Sin α-CSC α = 1Cos α-sec α = 1 Sin α/COS α = tan α = sec α/CSC αCos α/sin α = cot α = CSC α/sec α Sin2 α + cos2 α = 11 + tan2 α = sec2 α1 + cot2 α = csc2α Induction Formula

= Scanner.nextint ();inty = Scanner.nextint ();intmax = Scanner.nextint (); coordinate[] cos =NewCoordinate[n +1]; for(inti =0; I NewCoordinate (Scanner.nextint (), Scanner.nextint (), Scanner.nextint ()); }//Last person's coordinates as a 0-length bezelCos[n] =NewCoordinate (x, x, y); System.out.println (earliesttime (COS, max)); } scanner.close (); }/** * Minimum time for next 100 layers * * @p

Reciprocal Sin α * CSC α = 1 Cos α * sec α = 1 Tan α * cot α = 1 Commodity count Tan α = sin α/COS α Cot α = cos α/sin α 　 Sum of squares Sin2 α + cos2 α = 1 1 + tan2 α = sec2 α 1 + cot2 α = csc2α Two corners and Sin (α ± β) = sin α cos β +

the longitude and latitude of the second vertex B to (lonb, latb) based on the zero-degree longitude line, the longitude is a positive value (longpolling), the longitude is a negative value (-longpolling), and the latitude is 90-latitude ), when the latitude value is 90 + (90 + latitude), the two points after processing are counted as (mlona, mlata) and (mlonb, mlatb ). According to the triangle derivation, the following formula is obtained for calculating the distance between two points: C = s

This was discovered by a boy surnamed.Boy surface.This image is similar to a Rome image and is a three-point image. It can be said that it is made of a Rome surface change. This article will show the generation algorithms and cut diagrams of several boy surfaces, and generate mathematical graphics using script code with custom syntax. for related software, see: Mathematical graphics visualization tool. This software is free and open-source. QQ chat group: 367752815 In geometry,Boy's surfaceIs an

Function Getvis ( $pagesid, $retime) {Global $err, $conf, $HTTP _cookie_vars,$_cookie; if (isset ($_cookie[' ant ')) $cot =$_cookie[' ant '];ElseIf (isset ($HTTP _cookie_vars[' ant ')) $cot = $HTTP _cookie_vars[' ant '];else $cot = ';$cos =preg_split ("/x/", $cot);$max =sizeof ($cos);for ($c =0; $c if (strlen ($cos [$c]) ==10) {$id =substr ($

The idea of the C program is roughly to clarify, usually not programming, a lot of minor issues, but fortunately accompanied by a mother-in-law, quickly resolved. Draw a 16-point FFT. This is not an optimized program, but it should be the case for a clear idea. The next step is FPGA. Japanese words are not backed up yet. This week is messy, and every week is messy. Similar to the previous program, the four-point data input is divided into eight equations in the real imaginary part. About the ro

First, the cosine theorem is proved:Having an edge of a, B, c, the corresponding angle is a_angle, B _angle, c_angleThe following relationship can be found as a diagonal line from the point to the corresponding edge:(1)A = B * cos c_angle + c * cos B _angle(2)B = a * cos c_angle + c * cos a_angle(3)C = a *

springColumn CoordinateEquation: r = 5Theta = T * 3600.Z = (sin (3.5 * theta-90) 24 * tFigure 12. Linear line.Playing the cardEquation: A = 10X = 3 * a * t/(1 (t ^ 3 ))Y = 3 * a * (t ^ 2)/(1 (t ^ 3 ))Figure 23. helical curve)Cylindrical)Equation: r = TTheta = 10 t * (20*360)Z = T * 3Figure 34. Butterfly CurveSpherical CoordinatesEquation: FIG = 8 * tTheta = 360 * T * 4Phi =-360 * T * 8Figure 45. InvoluteCartesian coordinate systemEquation: r = 1Ang = 360 * tS = 2 * pI * r * tX0 = S *

; Center point float e[2]; Half-length, half-width}; Float Minarearec (point *pts, int ptsnum, OBB obb) {float Minarea = Flt_max; for (int i = 0, j = ptsNum-1; i Rotating caliper (rotating jam) algorithmUsing the rotary caliper algorithm can reduce the time consumption of calculating the minimum bounding rectangle of a convex polygon.Take the coordinates of the two points to form a parallel line, rotating two lines, when the line and polygon one edge coincident, the calculation of the

Solve $$\int_0^{\frac{\pi}{2}} {{{\ln}^2}\sin XDX} = \int_0^{\frac{\pi}{2}} {{{\ln}^2}\cos XDX} $$ value.Apparently \[\int_0^{\frac{\pi}{2}} {{{\ln}^2}\sin XDX} \underline{\underline {{\text{order}x = \frac{\pi}{2}-t}}} \int_0^{\frac{\ Pi}{2}} {{{\ln}^2}\cos xdx}.\]Remember to ask for integral for $i$, notice(1) $\int_0^\pi {{\ln}^2}\sin \left (x \right) dx} = 2\int_0^{\frac{\pi}{2}} {{{\ln}^2}\sin XDX} = 2

Not to mention nonsense: 0. Overview-Demand Analysis-function description-restrictions and defects improvement + knowledge preparation 1. essential differences between Taylor series and Fourier series, Taylor's expansion 2. Function projection and vector orthogonal 3. the derivation of the two invariant functions is e ^ X, SiNx, and cosx, Which is why Fourier conversion is required! 4. Fourier technology push process 5. Appendix references 0. In some cases, especially in image processing, the am

What is coordinate rotation? For example, after the (blue) Ball rotates a angle around a center point and reaches the position of the (red) Ball, the coordinates of the relative center of the red ball are: X1 = DX * Cos (a)-dy * sin () Y1 = Dy * Cos (A) + dx * sin () This is the coordinate rotation formula. If you want to reverse rotation, you need to modify the formula. There are two methods:

quantity product (magnitude), which represents the distance from the origin to the coordinate point, θ is the phase angle (phase angle), which represents the angle from the positive direction of the x-axis to a vector, and the following four formulas are the calculation methods:We can also convert polar to rectangular coordinates using the following equation:A + JB = M (cosθ+ j sinθθ) In this equation, the left side is a Cartesian expression, and the right is a polar coordinate expression.There

*pow_sign(sin(u), a)n = r*pow_sign(cos(u), b)y = n*cos(v)z = n*sin(v) vertices = D1:100 D2:100u = from 0 to (PI) D1v = from 0 to (2*PI) D2a = rand2(0.1, 4)b = rand2(0.1, 4)r = 10.0n = r*pow_sign(sin(u), a)y = r*pow_sign(cos(u), b)x = n*cos(v)z = n*sin(v) (2) hyper-sphere (thin) vertices = D1:100 D2:100u = from 0 to (2

a1, a2, a3, b1, b2, b3, c1, c2, c3;A1 = cos (r) * cos (t)-sin (r) * sin (s) * sin (t );A2 =-cos (r) * sin (t)-sin (r) * sin (s) * cos (t );A3 =-sin (r) * cos (s );B1 = cos (s) * sin (t );B2 =

the same time, so that J in the denominator can be eliminated.The plural also conforms to the exchange law, combination law, and allocation Law in the algebraic operation:A B = B(A + B) + C = a + (B + C)A (B + C) = AB + AC 2. Polar representation of the plural number As mentioned above, Cartesian coordinates are used to represent the plural. In fact, polar coordinates are more commonly used, for example: M in is the magnproduct, which indicates the distance from the origin point to the coordi

A few days ago, I published an article about how to generate a curve. For more information, see Mathematical graphics (1.11) Rose Line Mathematical graphics (1.27) flowers In this section, the two-dimensional curve is converted into a three-dimensional curve, which looks much more beautiful. For related software, see: Mathematical graphics visualization tool. Use script code with custom syntax to generate mathematical graphics. (1) rose Wire vertices = D1:4000 D2:6n = 8u = from 0 to (n*PI) D1v

1 ($4 \ times 10' = 40' $) (1) $ \ DPS {\ lim _ {x \ to 0} \ frac {\ SiN x-\ arctan x }{\ Tan x-\ arcsin x }}$. Answer: $ \ beex \ Bea \ lim _ {x \ to 0} \ frac {\ SiN x-\ arctan x} {\ Tan x-\ arcsin x} = \ lim _ {x \ to 0} \ frac {\ SEZ {X-\ frac {x ^ 3} {6} + O (x ^ 3 )} -\ SEZ {X-\ frac {x ^ 3} {3} + O (x ^ 3 )}} {\ SEZ {x + \ frac {x ^ 3} {3} + O (x ^ 3 )} -\ SEZ {x + \ frac {x ^ 3} {6} + O (x ^ 3 )}} \ =\ frac {-\ frac {1} {6} + \ frac {1} {3 }{\ frac {1} {3}-\ frac {1} {6 }\\ = 1. \ EEA