x=a(θ-sinθ) y=a(1-cosθ) X = a (θ - sin θ), y = a (1-cos θ). Find d ^ 2Y / DX ^ 2. Dy / DX = cot θ / 2, d ^ 2Y / DX ^ 2 = D / DX * dy / DX = D (dy / DX) / D θ * D θ / DX = D (dy / DX) / D θ * 1 / DX / D θ = I want to ask ha d ^ 2Y / DX ^ 2 how to solve d (dy / DX) / D θ * D θ / DX = D (dy / DX) / D θ * 1 / DX / D θ Z

x=a(θ-sinθ) y=a(1-cosθ) X = a (θ - sin θ), y = a (1-cos θ). Find d ^ 2Y / DX ^ 2. Dy / DX = cot θ / 2, d ^ 2Y / DX ^ 2 = D / DX * dy / DX = D (dy / DX) / D θ * D θ / DX = D (dy / DX) / D θ * 1 / DX / D θ = I want to ask ha d ^ 2Y / DX ^ 2 how to solve d (dy / DX) / D θ * D θ / DX = D (dy / DX) / D θ * 1 / DX / D θ Z

Higher mathematics volume 1 (Tongji Edition), Chapter 2, there is a question: Children's shoes. I wrote the steps above. That is to ask the meaning. Answer: if the independent variable of θ is changed to x, we must ask DX / D θ, because the second derivative is y to x, not to θ, so we need to replace it

cos(x)=1 +sin(x)cos(x)
Using two methods to find x

1. We use sin (x) 2 + cos (x) 2 = 1 to replace sin (x), and then solve the problem
2. Using the formula of double angle on both sides at the same time, we can get: 2cos (2x) = 1 + sin (2x), through the transformation that the square of (cos2x) is equal to the square of 1-sin (2x), we can get the result finally

When t takes all the values, what is the area of the figure enclosed by the straight line X * cos (T) + y * sin (T) = 4 + sin (T + 45 °) * 2 ^ (1 / 2)

16π

Excuse me: given the equation of a circle is p = cos θ - sin θ, find the center and radius of the circle

The formula P = cos θ - sin θ multiply P on both sides,
P^2=Pcosθ-Psinθ
Because p ^ 2 = x ^ 2 + y ^ 2
X=Pcosθ
Y=Psinθ
be
X^2+Y^2=X-Y
Namely
(X-1/2)^2+(Y+1/2)^2=1/2
Then the center of the circle is (1 / 2, - 1 / 2) and the radius is 1 / 2

The equation of the circle with the smallest area and the intersection of the line 2x + y + 4 = 0 and the circle (x + 1) ^ 2 + (Y-2) ^ 2 = 4 is obtained

The circle whose diameter is the chord of intersection has the smallest area
Substituting 2x + y + 4 = 0 into (x + 1) ^ 2 + (Y-2) ^ 2 = 4
The result is: 5x ^ 2 + 26x + 33 = 0
Midpoint x = - 13 / 5, y = 6 / 5
The square of radius can be obtained by chord center distance: 4 / 5
The equation is: (x + 13 / 5) ^ 2 + (y-6 / 5) ^ 2 = 4 / 5

The equation of circle is x ^ 2 + (Y - (1 / 2) a) ^ 2 = (1 / 4) a ^ 2. How to express it as x =, y = (including sin, COS)

4x²/a²+4（y-0.5a）²/a²=1
4x²/a²=sin²
4（y-0.5a）²/a²=cos²
∴x=0.5asin
y=0.5acos+0.5a

Solving the equation of the smallest circle with the intersection of two circles x ^ 2 + y ^ 2 = 5 and (x-1) ^ 2 + (Y-1) ^ 2 = 16

x²+y²=5
The center of the circle is (0,0)
(x-1)²+(y-1)²=16
The center of the circle is (1,1)
Center distance = 2 under root
The chord length of the intersection of two circles = 2 × 2 under the root sign △ 2 = 2 under the root sign
When the chord length = the diameter of the circle, the area of the circle is the smallest
The middle point of the center of two circles is the center of the circle (1 / 2,1 / 2)
So the circular equation is (x-1 / 2) &# 178; + (Y-1 / 2) &# 178; = 9 / 2

Verification: cos & # 178; X + cos & # 178; (x + a) - 2cosxcosacos (x + a) = Sin & # 179; a

Left = (cosx) ^ 2 + [cos (x + a)] ^ 2 - [cos (x + a) + cos (x-a)] cos (x + a)
=(cosx)^2-cos(x-a)cos(x+a)
=(1/2)(1+cos2x-cos2x-cos2a)
=(1/2)(1-cos2a)
=(sina)^2,
The number on the right is not right

How to calculate cos (α - 35 °) cos (α + 25 °) + sin (α - 35 °) sin (α + 25 °)

The original formula = cos [(α - 35 °) - (α + 25 °)]
= cos (-60°)
= cos 60°
= 1/2.
= = = = = = = = =
cos (A-B) =cos A cos B +sin A sin B,
cos (-A) =cos A.

Given that π / 4 ＜ x ＜ π / 2, SiNx cosx = 1 / 5, evaluation: (1) SiNx + cosx; (2) 3sin ^ 2x + cos ^ 2x-4sinxcosx

1):sinx-cosx=1/5,sinxcosx=12/25(sinx+cosx)^2=(sinx-cosx)^2+4sinxcosx=1/25+48/25=49/25sinx+cosx=7/52):3sin^2x+cos^2x-4sinxcosx=1+2sinx(sinx-2cosx)=1+8/5*(4/5-6/5)=9/25