In polar coordinates, the equation of a circle with (A2, π 2) as its center and A2 as its radius is______ .

In polar coordinates, the equation of a circle with (A2, π 2) as its center and A2 as its radius is______ .

As shown in the figure, ∵ apo is the circumference angle of the diameter Ao of ⊙ o, ∵ apo = π 2. ∵ ρ = ACOS (π 2 − θ) = asin θ. ∵ ρ = asin θ. So the answer is: ρ = asin θ
How to write the polar equation of a circle with the center of a (1, π / 4) and radius of 1?
The general polar equation of a circle is
p^2=2pmcos(&-n)+m^2=r^2
Center (m, n), radius r
Just substitute it directly
The equation is finally
p^2-2pcos(&-π/4)=0
x=sina+1
y=cosa+π/4
The polar coordinate equation of a circle with a center of (a, π 2) and radius of a & nbsp; is ()
A. ρ=acosθB. ρ=2acosθC. ρ=asinθD. ρ=2asinθ
As shown in the figure, ∠ OQP = θ, ∠ QPO = 90 °, ρ = 2asin θ
If we know that the center of the circle is a (1, π / 4) and the radius is 1, we can solve the polar coordinate equation
The general polar equation of a circle is
p^2=2pmcos(&-n)+m^2=r^2
Center (m, n), radius r
Just substitute it directly
The final equation is
p^2-2pcos(&-π/4)=0
In the triangle ABC, the opposite sides of the angle ABC are ABC. Given that B = 12 π, C = B (1 + 2cosa), find the angle A
In triangle ABC, there is a sine theorem
B / SINB = C / sinc, i.e. B / C = SINB / sinc
And because:
C = B (1 + 2cosa) so: B / C = 1 / (1 + 2cosa)
So: SINB / sinc = 1 / (1 + 2cosa) (1)
From the sum of the inner angles of the triangle, we know that a + B + C = π, so C = π - (a + b), and substitute (1)
SINB / sin [π - (a + b)] = 1 / (1 + 2cosa) reduction:
sinB+2sinBcosA=sin(A+B)=sinAcosB+cosAsinB
sinB=sinAcosB-cosAsinB=sin(A-B)
B = A-B or π - B = a-b
So: a = π / 6 or a = π, because a
C = B (1 + 2cosa), so cosa = 1 / 2 (C / B - 1), so a = arccos [1 / 2 (C / B - 1)]...
If the inequality m (x-1) about X is greater than the solution set of X & # - x (x / 1 < x < 2), then the value of real number m is
Remember this topic:
x=1,x=2
It must be: m (x-1) = x & # 178; - the two roots of X,
therefore
What we need to do is to substitute x = 1 and x = 2 (generally, we only need to substitute one)
This problem needs to be replaced by x = 2
m=2²-2=2
M=2
Analysis: according to the relationship between the solution set of quadratic inequality with one variable and the root of the corresponding quadratic equation, it is judged that x = 2 is the root of the equation, and the value of M is obtained by substituting x = 2
The solution set of ∵ m (x-1) > x2-x is {x | 1 ＜ x ＜ 2},
1,2 are the two roots of the equation m (x-1) = x2-x
Substituting x = 2
So the answer is: 2
Comments: the solution set problem of quadratic inequality with one variable is often transformed into the solution problem of corresponding quadratic equation with one variable
Analysis: according to the relationship between the solution set of quadratic inequality with one variable and the root of the corresponding quadratic equation, it is judged that x = 2 is the root of the equation, and the value of M is obtained by substituting x = 2
The solution set of ∵ m (x-1) > x2-x is {x | 1 ＜ x ＜ 2},
1,2 are the two roots of the equation m (x-1) = x2-x
Substituting x = 2
So the answer is 2
Comments: solving the solution set problem of quadratic inequality of one variable is often transformed into the solution problem of corresponding quadratic equation of one variable
Given the function f (x) = cos ^ 2x + SiNx + A-1, if f (x) = 0, there is a real number solution to find the value range of A
The square of cosx
Cos ^ 2x + SiNx + A-1 = 0 has real number solution, which is equivalent to a = 1 - cos ^ 2x - SiNx has real number solution, that is, the value of a is in the range of function g (x) = 1 - cos ^ 2x - SiNx. G (x) = 1 - cos ^ 2x - SiNx = sin ^ 2x + SiNx = (SiNx + 1 / 2) ^ 2-1 / 4, SiNx = - 1 / 2, G (x) takes the minimum value - 1 / 4. When SiNx = 1, G (x) takes
The known sequence {an} satisfies the general formula an = (1 + 2 + 3 +...) +n) / N, BN = 1 / (an * an + 1), find the first n terms and Sn of sequence {BN}
Because 1 + 2 + 3 + +N = (n + 1) n / 2, so an = (n + 1) / 2, so BN = 4 / ((n + 1) (n + 2)) = 4 (1 / (n + 1) - 1 / (n + 2)) = = > Sn = 4 (1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 +...) +1 / (n + 1) - 1 / (n + 2)) = 2n / (n + 2)
8sin ^ 2 (B + C) / 2-2cos2a = 7 for angle a
B + C = π - a8sin ^ 2 (B + C) / 2-2cos2a = 4 (1-cos (B + C)) - 2cos2a = 4 (1-cos (π - a)) - 2cos2a = 4 + 4cosa-2 (2cos ^ 2a-1) = - 4cos ^ 2A + 4cosa + 6 = 7  4cos ^ 2a-4cosa + 1 = 0 (2cosa-1) ^ 2 = O cosa = 0.5 in the triangle, a is 60 degrees
Given the set M = {y y = x ^ 2-4x + 3, X belongs to real number}, and the set n = {y y = - x ^ + 2x + 8, X belongs to real number}, find the intersection n of M
It's better not to reprint it on the Internet. I want details. The more detailed, the better
thank you
y = x² - 4x + 3
= x² - 4x + 4 - 1
= (x - 2)² - 1
≥ -1
y = -x² + 2x + 8
= -x² + 2x - 1 + 9
= -(x - 1)² + 9
≤ 9
So m ∩ n = {y | - 1 ≤ y ≤ 9}