Let Tan (θ / 2) = t, prove: sin θ = 2T / (1 + T ^ 2), cos θ = (1-T ^ 2) / (1 + T ^ 2), Tan θ = 2T / (11T ^ 2) Let Tan (θ / 2) = t, prove: sin θ = 2T / (1 + T ^ 2), cos θ = (1-T ^ 2) / (1 + T ^ 2), Tan θ = 2T / (1-T ^ 2)

Let Tan (θ / 2) = t, prove: sin θ = 2T / (1 + T ^ 2), cos θ = (1-T ^ 2) / (1 + T ^ 2), Tan θ = 2T / (11T ^ 2) Let Tan (θ / 2) = t, prove: sin θ = 2T / (1 + T ^ 2), cos θ = (1-T ^ 2) / (1 + T ^ 2), Tan θ = 2T / (1-T ^ 2)

sinθ=2sinθ/2*cosθ/2
=2tanθ/2*cosθ/2*cosθ/2
=2tanθ/2(cosθ/2)^2
=2t*(cosθ/2)^2
And (COS θ / 2) ^ 2
=1/（secθ/2)^2
=1/[1+（tanθ/2)^2]
=1/1+t^2
So sin θ = 2T / (1 + T ^ 2)
In the same way
cosθ=2(cosθ/2)^2-1
=2/1+t^2-1
=(1-t^2)/(1+t^2)
tanθ=sinθ/cosθ=2t/(1-t^2)

Solving a problem of definite integral for volume of revolution
In the book, I see the formula: DV = π [(x + DX) ^ 2-x ^ 2] f (x)
It is concluded that v = 2 π∫ XF (x) DX~
Why?
Is the square of DX equal to 0?

The smaller DX is, the more accurate the formula is, so DX is very small, so the square is smaller, which can be considered as 0
Never forget that integrals are based on infinite approximation
Good luck!

Ask a question for the students about using definite integral to calculate the volume of the body of revolution
The proof of question 19 in exercise 6-2 in Volume I of the fifth edition of Tongji University roughly means to prove the volume of the body of revolution formed by the rotation of a plane figure around the Y axis with the infinitesimal method
We know the equation curve about X, x = a, x = B, and the volume of trapezoid with curved edge surrounded by X axis rotating around Y axis (please see these XY)
V=2pi∫(b,a) x f(x)dx
This volume element is 2pi XF (x) DX. How do you get it?
More details, thank you

Taking a small section of the curve f (x) and projecting it to the x-axis, the trapezoid with a small curved edge can be approximately regarded as a rectangle with a height of F (x) and a width of DX from X to x + DX. If the trapezoid with a small curved edge is rotated around the y-axis, the resulting rotator can be regarded as a cylinder with a circular bottom (the radius of the small circle is x, the radius of the big circle is x + DX) and a height of F (x)

Definite integral for area and volume of revolution,
 

The plane figure D is surrounded by parabola y = 1-x ^ 2 and X axis, and the volume of the body of revolution obtained by D rotating around X axis

∫ π (1-x ^ 2) ^ 2DX integral interval [0,1]
=π(x+x^5/5-2x^3/3) [0,1]
Substituting the upper and lower limits of integral
We get 8 π / 15

Given that the function y = ax3-15x2 + 36x-24 has an extreme value at x = 3, the decreasing interval of the function is ()
A. （-∞，1），（5，+∞）B. （1，5）C. （2，3）D. （-∞，2），（3，+∞）

For the derivative of the function y = ax3-15x2 + 36x-24, we get y '= 3ax2-30x + 36 ∵ the function y = ax3-15x2 + 36x-24 has an extreme value at x = 3, when x = 3, y' = 27a-54 = 0, the solution is a = 2, so we can get the analytic formula of the function y = 2x3-15x2 + 36x-24, y '= 6x2-30x + 36, and solve the inequality y' < 0, 2 < x < 3 ∵

Find the maximum value of the function f (x) = 2x ^ 3-15x ^ 2 + 36x-24 in the interval [1,13 / 4]

The solution of the maximum value of cubic function is: first, the derivative is obtained, and the abscissa of the extreme point is obtained. Then, the extreme point is compared with the function value of the upper and lower bounds of the domain
f'(x)=6x^2-30x+36=6(x-2)(x-3)
So we have minimax when x = 2, x = 3
f(2)=4,f(3)=3
And f (1) = - 1, f (13 / 4) = 103 / 32
So the maximum is 4 and the minimum is - 1

Given that the function f (x) = - 13x3 + bx2-3a2x (a ≠ 0) obtains the extreme value at x = a, (1) use x, a to represent f (x); (2) let the function g (x) = 2x3-3af '(x) - 6a3, if G (x) has a minimum value in the interval (0, 1), find the value range of real number a

(1) We get f '(x) = - x2 + 2bx-3a2, because f' (a) = 0 {B = 2A} f (x) = - 13x3 + 2ax2-3a2x, so f (x) = - 13x3 + 2ax2-3a2x. (2) as we know, G (x) = 2x3 + 3ax2-12a2x + 3a3, let g '(x) = 0 {x = a or x = - 2A, if a ＞ 0} when x ＜ a or X ＞ - 2A, G ′

What is the formula of sector area in polar coordinates

The vertex of the sector is the pole and one side is the polar axis
Let: the sector apex angle be θ (radian), and the radius be r
Then the sector area s = (1 / 2) θ R & sup2

Solution: formula of side area in polar coordinates
rt

The radius of the sector is the generatrix of the cone, and the arc length of the sector is the circumference of the bottom of the cone. We know that the area formula of the sector is: S = 1 / 2lr, that is, the area of the sector is equal to half of the arc length multiplied by the radius. For example, OA is the radius r, so the arc length of the sector is equal to 2 π R, SA is the radius L, so