If the surface area of the cone is 15 π and the center angle of the side view is 60 °, then the volume of the cone is 15 π___ ．

If the surface area of the cone is 15 π and the center angle of the side view is 60 °, then the volume of the cone is 15 π___ ．

Let the bottom radius of the cone be r, and the generatrix be l, then 2 π r = 13 π L, we get l = 6R, s = π R2 + π R · 6R = 7 π R2 = 15 π, we get r = 157, the height of the cone H = l2-r2 = 35r, that is, H = 35.157, v = 13 π r2h = 13 π × 157 × 35 × 157 = 2537 π

If the axial section of a cone is an equilateral triangle and its area is root 3, then the generatrix length of the cone is

Because the generatrix length is the side length of the triangle, a = 2 can be obtained from S = root 3 / 4 A ^ 2

Given x ∈ [0, π / 2], find the range of the function g (x) = cos (π / 12-x) - cos (5 π / 12 + x)

G (x) = cos (π / 12-x) - cos (5 π / 12 + x) = sin (5 π / 12 + x) - cos (5 π / 12 + x) = √ 2 * [√ 2 / 2 * sin (5 π / 12 + x) - √ 2 / 2 * cos (5 π / 12 + x)] = √ 2 * sin (5 π / 12 + X - π / 4) = √ 2 * sin (x + π / 6) because x ∈ [0, π / 2], so x + π / 6 ∈ [π / 6,2 π / 3], so sin (x + π / 6)

The function f (x) = cos (x + PAI) / 4) (x belongs to ["PAI) / 12,5" PAI) / 6] in the range of

If x is in [π / 12,5 π / 6], then x + π / 4 is in [π / 3,13 π / 12]
When x = π 3 / 4, there is a minimum value of - 1, and when x = π / 12, there is a maximum value of 1 / 2

Find the function y = cos (x + π / 3), X belongs to the range of (0, π / 3)

[-1/2,1/2]

The range of function f (x) = (2 + cosx) / (2-cosx) is?
If the minimum value of function y = Xinx ^ 2 + 2cosx in the interval [- 2 / 3pai, a] is - 1 / 4, then the value range of a is?
Negative two thirds

(1) Let cosx = t, t ∈ [- 1,1], then f (x) = (2 + T) / (2-T), then f '(x) = 4 / (2-T) ^ 2 > 0, so f (x) is an increasing function on R, so f (x) min = f (- 1) = 1 / 3, f (x) max = f (1) = 3, so the range of F (x) is [1 / 3,3] (2) y = SiNx ^ 2 + 2cosx = 1-cosx ^ 2 + 2cosx = - (cosx-1) ^ 2 + 2, because in [- 2 π /

Find the range of function f (x) = 5sinx / (cosx-5)
【f(x)=5(sinx-0)/(cosx-5)
It is five times the slope of the line between point a (cosx, SiNx) and point B (5, 0)
Where a is on the circle x ^ 2 + y ^ 2 = 1
The easy to know value range is (- 5 / radical 24,5 / radical 24)]
There is no need for such a process~

From F (x) = 5sinx / (cosx-5), f (x) (cosx-5) = 5sinx
5sinx-f (x) cosx = - 5F (x)
sin[x-arctan（f(x)/5）]=-5f(x)/√[25+f(x)^2],
Then | - 5F (x) / √ [25 + F (x) ^ 2] | ≤ 1,
25*f(x)^2≤25+f(x)^2,f(x)^2≤25/24,
-5√6/12≤f(x)≤5√6/12
That is to say, the range of function f (x) is [- 5 √ 6 / 12,5 √ 6 / 12]

Find the range of y = cos2x-4sinx

y=cos2x-4sinx
=1-2sin²x-4sinx
=-2(sin²x+2sinx)+1
=-2(sinx+1)²+3
When SiNx = - 1, y has a maximum value of 3
When SiNx = 1, y has a minimum value = - 8 + 3 = - 5
Therefore, the range of the function is y ∈ [- 5,3]

The range of the function y = (cosx · sin ^ 2) / 1-cosx is
The range of function y = (cosx · sin ^ 2) / 1-cosx is
Wait for the answer online!
There should be a specific process, thank you

sin^2x+cos^2x=1
So y = cosx * (1-cos ^ 2x) / (1-cosx)
=cosx*(1+cosx)(1-cosx)/(1-cosx)
=cosx(1+cosx)
=cos^2x+cosx
=(cosx+1/2)^2-1/4
The denominator is not equal to 0
Cosx is not equal to 1
So - 1

Given that the domain of F (x) = loga (x-ka) is a, and the domain of G (x) = loga (x ^ 2-A ^ 2) is B, find a ∩ B

To make f (x) = 2loga (x-ka), G (x) loga (x ^ 2-A ^ 2) meaningful,
Then there must be x-ka > 0, x ^ 2-A ^ 2 > 0
When a = {x > Ka}, B = {, x > A or x1, Ka > a > 0, a intersects B = {x > Ka}
0