-Why is 2 sin 2T / cos t equal to - 4 sin t

-Why is 2 sin 2T / cos t equal to - 4 sin t

-2 sin 2t / cos t=sintcost*sintcost
sintcost=sin2t/2
Double angle formula
If you are satisfied, please click [satisfied] in the upper right corner~

What is the relationship between traction and weight power?
400kg car, 1020w mechanical energy, not resistance, how to calculate traction and speed? (running on flat ground)
I'm dizzy. I'll give it to whoever can make me understand-

Power = speed × traction, which is the relationship between the three. And from this, we can see the resistance when the maximum power = maximum speed × maximum speed. At the same time, we find a problem. If the resistance is not included, the speed will increase endlessly. If the answer is still not satisfied, maybe your problem is too simple, and the specific problem can be analyzed concretely

The relationship between the speed V (M / s) and the tractive force F (n) of a certain car with a certain power is given, and the function analytical expression between V and F is written
When the traction force is 1200n, what is the speed of the car? When the speed is less than 30m / s, what is the range of the traction force F

P=W/t=（f*S）/t=f*(S/t)=f*V
So f (n) = P / v
What is the power
Then it can be calculated

Is the traction equal to the resistance? In the process of getting on the bridge, because P = FS, the power of the car is fixed, so the slower the speed of the car, the greater the traction?
Our teacher said that the slower the car is, the greater the traction. She said that the car will slide when the speed is fast, but in real life, the car will slide when the speed is slow
Wrong number, P = FV
Downstairs, I asked, is traction equal to resistance?

You remember wrong? It should be p = Fv (F is traction, V is speed). When the power is constant, the smaller V is, the larger f is
PS: in real life, the reason why the speed is slow is that the power of the car is not fixed

Examples of calculating definite integral with double integral
For example, classical integrals like e ^ (- x ^ 2) DX,
I've also seen one:
Arctan (x) / (x * (1-x ^ 2) ^ 0.5) DX is also calculated by double integral
Ask the master to give some examples

For example, SiNx / X can be done by double integral, and I just did the following:
Here's an idea:

Find a tangent line of the curve y = LNX (2 ≤ x ≤ 6) to minimize the area of the figure enclosed by the tangent line, the straight line x = 2, x = 6 and the curve y = LNX

The slope of y = LNX at point (U, LNU) is y '= 1 / x = 1 / u
Tangent equation y-lnu = 1 / u (x-u) y = x / U-1 + LNU 2

A definite integral problem for higher numbers
Let f (x) have a first order continuous derivative, and f (0) = 0. F '(0) is not equal to 0. Then the limit x tends to be 0 ‖ 0 to X ＾ 2 (f (T) DT) / (x ＾ 2 ‖ 0 to XF (T) DT)
∥ is the integral symbol

=lim(2xf(x^2))/(2x∫(0,x)f(t)dt+x^2f(x))
=lim(2f(x^2))/(2∫(0,x)f(t)dt+xf(x))
=lim(4xf'(x^2))/(3f(x)+xf'(x))
If f (0) = 0. F '(0) is not equal to 0, there is no way to do it. The limit of numerator and denominator is 0

The number of intersections between the straight line 3x + 2Y = a (a > 0) and the curve {x = ACOS β, y = asin β (β is a parameter) is?

By title
x²+y²=(acosβ)²+(asinβ)²=a²
That is X & sup2; + Y & sup2; = A & sup2;
The distance between the center of circle X & sup2; + Y & sup2; = A & sup2; and the line 3x + 2Y = a
d=|-a|/√(3²+2²)=a/√13

Find the area a of the figure enclosed by the ellipse x = ACOS θ, y = asin θ

It should be: circle x = ACOS θ, y = asin θ enclosed by the area of a figure
The equation of a circle is x ^ 2 + y ^ 2 = a ^ 2
If radius is a, then area a = π a ^ 2

The derivative of ∫ O-X f (T) DT is f (x). How to calculate the derivative of ∫ 0-x (x-t) f (T) DT

First, the original form is divided into several parts
The original formula = ∫ (0 ~ x) XF (T) DT - ∫ (0 ~ x) TF (T) DT ①
For the integral on the left of the above formula, X is a constant in the integrand,
So it is equal to X ∫ (0 ~ x) f (T) DT
Therefore, the derivation of formula (1) can be obtained according to the previous method
(x)'∫(0~x)f(t)dt+x[∫(0~x)f(t)dt]'-[∫(0~x)tf(t)dt]'
=∫(0~x)f(t)dt+xf(x)-xf(x)
=∫(0~x)f(t)dt