tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3t+a)cos^2(2/5π+2)=-2/1 Seeking cosa * | Tana|

tan(a-4π)cos(π+a)sin^2(a+3π)/tan(3t+a)cos^2(2/5π+2)=-2/1 Seeking cosa * | Tana|

Tan (A-4 π) = - tanacos (π + a) = - cosasin ^ 2 (a + 3 π) = sin ^ 2atan (3 π + a) = - tanacos ^ 2 (5 π / 2 + a) = sin ^ 2A equation left = - Cosa ﹤ cosa = 1 / 2 | Tana | = root 3, so the answer = two thirds root 3. This friend has too many typing errors. 5 / 2 is written as 2 / 5, 1 / 2 is written as 2 / 1. π is written as T. finally

Derivation of curvature formula
Step 1: because Tan α = y ', so sec & # 178; α (D α / DX) = y ″
Where does (D α / DX) come from?

Y & # 39; = Tan α & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; / /: & nbsp; Y & # 39; is the tangent value of the angle between the tangent line and the x-axis at the X point; Y & # 39; = [D (Tan α) / D α] (D α / DX) & nbsp; & nbsp; = sec & # 178; α & nbsp; (D α / DX) & nbsp; & nbsp; &

Polar coordinate equation of circle! The formula is not clear! Find the direction~
In the polar coordinate system, the equation of the circle whose center is at (R0, φ) and radius is a is
r^2 - 2 r r_ 0 \cos(\theta - \varphi) + r_ 0^2 = a^2
I don't know where r comes from, where theta comes from, and how it comes out

I don't quite understand your question, but in general, the polar coordinate system equation of a circle can be transformed into the rectangular coordinate system equation of a circle
X=pcosθ,Y=psinθ
The rectangular coordinate system equation of circle is (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2
Where a is the distance from the center of the circle to the X axis, B is the distance from the center of the circle to the Y axis, and R is the radius of the circle

How to calculate the curvature coefficient

The curvature coefficient represents the degree of curvature of an arc at a point. The more curved the arc is, the greater the curvature coefficient is
That is to say, when the radius of the circle corresponding to an arc is smaller, the curvature coefficient is larger;
The larger the radius of the circle corresponding to the arc, the smaller the curvature coefficient (this should not be difficult to understand,
Curvature coefficient = 1 / radius of arc

How to calculate the curvature of a circle in high numbers

In the same way, according to the curvature calculation formula, first write out the function of the semicircle, and find the first derivative and the second derivative respectively, then the curvature is
K = | y '' / (1 + y '^ 2) ^ 3 / 2 | the radius of curvature is the reciprocal of the curvature. In fact, the radius of curvature is the radius of the circle

Radius of curvature circle
Y = x ^ 2, find the radius of curvature at x = 1

y=x^2,y'=2x,y''=2:
K=|y"|/[(1+y'^2)^3/2]
Radius of curvature R = 1 / k = (radical 5) * (5 / 2)

Find the linear equation of the common chord of circle x ^ 2 + y ^ 2-4 = 0 and circle x ^ 2 + y ^ 2-4x-12 = 0?

The line equation where the common string is located is the line equation where they have something in common, so we only need to transform the two formulas into a straight line. Subtracting the two formulas and eliminating the quadratic term is the line equation, which is the line equation where the common string is located. When x ^ 2 + y ^ 2-4 = x ^ 2 + y ^ 2-4x-12 = 0, subtracting is: 4x + 12 = 4x + 2 = x = - 2, y = 0

The tangent equation of curve f (x) = LNX / X passing through point P (1,0) is

(1,0) may not be the tangent point, so we can't bring the point directly
Let y = kx-k. in conjunction with the original formula, K is obtained
By deriving f (x), f (x) = (1-inx) / X2 is obtained
I got X,
Get the equation

Definite integral volume of revolution
The volume formed by y = SiNx (0 ≤ x ≤ π) rotating around the y-axis is integrated as v = π∫ (lower limit 0, upper limit 1) [(π - arcsiny) ^ 2 - (arcsiny) ^ 2] dy
How do you get this integral,

Finding the volume of a rotating body (definite integral)
An arch of cycloid x = a (t-sint), y = a (1-cost), y = 0, around a straight line y = 2A
I can't draw a figure... I can't do it at all

The rotation axis y = 2A is just at the top of cycloid, and the volume of rotation body: v = ∫ π [4A & # 178; - (2a-y) & # 178;] DX, the integral interval of X is an arch [0,2 π a];
In the parametric equation, v = 8 π & # 178; a & # 179; - ∫ π (2a-a + acost) &# 178; * a (1-cost) DT, t = [0,2 π];
V=8π²a³-πa³∫(1+cost)²(1-cost)dt=8π²a³-πa³∫(1+cost)sin²t dt
=8π²a³-πa³∫sin²t dt=8π²a³-πa³∫(1-cos2t)dt/2=8π²a³-πa³/2；