The derivation process of gold substitution in senior high school physics
For an object on the earth's surface (if the earth's rotation is ignored) or a satellite moves around the earth in a circle, gravity is equal to universal gravitation, that is, Mg = (GMM) / R ^ 2, that is, GM = GR ^ 2
Physical gold substitution What are the requirements of gold substitution for objects? Should the object be on the surface? Must it be stationary? Is it relative static with the earth? Or is it possible to give a detailed scope of application?
The gold substitution formula GM = Gr? 2 is based on the fact that the universal gravitation of the object on the surface of the planet is approximately equal to gravity, that is, GMM / R? = mg
Therefore, the object must be on the earth's surface, but not necessarily stationary
Derivation of trigonometric function formula SiNx sinx0 = 2Sin (x-x0) / 2 * cos (x + x0) / 2
Using the sine formula of sum and difference of two angles: sin (θ ± γ) = sin θ cos γ ± cos θ sin γ. Sin α - sin β = sin [(α + β) / 2 + (α - β) / 2] - sin [(α + β) / 2 - (α - β) / 2] = sin [(α + β) / 2] cos [(α - β) / 2] + cos [(α + β) / 2] sin (α - β) / 2] - sin [(α + β) / 2] cos [(...)
∵ SiNx + sin (x + 2) = 2Sin (x + 1) cos = sin (x + 1), why this step?
The sum difference product formula is used
sinθ+sinφ=2sin[(θ+φ)/2]cos[(θ-φ)/2]
How does SiNx sin Π / 4 change to [2Sin (x - Π / 4) / 2] [cos (x + Π / 4) / 2]
This is the sum difference product formula. You can look at this
How is this equation transformed into △ y = sin (x + △ x) - SiNx = 2Sin (△ X / 2) cos (x + △ X / 2)? I don't know which formula of trigonometric function is used~
Sum difference product
sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]
sinα-sinβ=2cos[(α+β)/2]sin[(α-β)/2]
cosα+cosβ=2cos[(α+β)/2]cos[(α-β)/2]
cosα-cosβ=-2sin[(α+β)/2]sin[(α-β)/2]
Find the indefinite integral of ∫ [arctan √ X / √ (1 + x)] DX. √ denotes the root sign,
T = arctan √ x, sect = √ (1 + x), x = tan? T, DX = 2 Tan t * sec? T DT original formula = ∫ 2 T d (sect) = 2 T * sect - 2 ∫ sect DT = 2 T * sect - 2 ln | sect + tant | + C = 2 √ (1 + x) arctan √ X - 2 ln | (1 + x) + √ X
Indefinite integral arctan radical x DX
Step by step integration method
The original formula = xarctan √ X - ∫ xdarctan √ x
=xarctan√x-∫x/(1+x)dx
=xarctan√x-∫(x+1-1)/(1+x)dx
=xarctan√x-∫[1-1/(1+x)]dx
=xarctan√x-x+ln(1+x)+C
What is the indefinite integral of 1 + x ^ 2 under the root sign depends on the process or method
By using the second integral transformation method, let x = TANU, u ∈ (- π / 2, π / 2), then ∫ √ (1 + X?) DX = ∫ sec ∫ UDU = ∫ secudtanu = secutanu - ∫ tanudsecu = secutanu - ∫ tan? Usecudu = secutanu - ∫ sec? UDU + ∫ secudu = secutanu + 1 / 2ln | secu + Tan
Indefinite integral of (1-x ^ 2) / x ^ 2 under radical sign The denominator x ^ 2 is not in the root sign~
Let x = sin α, then DX = cos α * D α
∫√(1-x^2) *dx /x^2
=∫cosα * (cosα *dα) /(sinα)^2
=∫(cotα)^2 dα
=∫[(cscα)^2 -1] *dα
=∫(cscα)^2*dα - ∫ dα
=-cotα - α + C
=-√(1-x^2)/x - arcsinα + C