1 if the origin is outside the circle x ^ 2 + y ^ 2 + 2x + 4y-a = 0, then the value range of a is 0______ What is the maximum distance between the point on the circle (x-1) ^ 2 + (Y-2) ^ 2 = 9 and the line 3x + 4y-15 = 0______ What is the positional relationship between the line x + y + 3 = 0 and the circle (x-3) ^ 2 + (y + 3) ^ 2 = 5_____

1 if the origin is outside the circle x ^ 2 + y ^ 2 + 2x + 4y-a = 0, then the value range of a is 0______ What is the maximum distance between the point on the circle (x-1) ^ 2 + (Y-2) ^ 2 = 9 and the line 3x + 4y-15 = 0______ What is the positional relationship between the line x + y + 3 = 0 and the circle (x-3) ^ 2 + (y + 3) ^ 2 = 5_____

3 intersection

If the distance between any point on the circle (x-1) ^ 2 + (Y-1) ^ 2 = R ^ 2 (r > 0) and the origin is 1, the value range of R can be obtained

Connect (1,1) and (0,0)
The distance between the origin and the center of the circle is √ 2,
Then the distance from the point closest to the origin on the circle to the origin is | √ 2-r|
If there is always a point with a distance of 1 from the origin,
Need | √ 2-r|

Given that there are only two points on the circle C: (x-4) ^ 2 + (y-4) ^ 2 = R ^ 2 whose distance from the origin is 1, then the value range of R is

The distance from the origin to the center of the circle is d = 4 √ 2
The point whose distance from the center of the circle is equal to 1 is on a circle (radius is 1)
Two circles intersect
∴ r-1

Simplification ~ · cos (2x + π / 3) + SiNx ^ 2

The original formula is cos (2x + π / 3) + (1-cos2x) / 2
=1/2cos2x-√3/2sin2x+1/2-1/2cos2x
=-√3/2sin2x+1/2

How to reduce 1-cos (SiNx) to SiNx square / 2? Function and limit

The simplification of COS (2x + π / 3) + SiNx

The original formula = cos2xcos3: π - sin2xsin3: π + sin2x = 2: 1cos2x-2: root sign 3sin3: π + sin2x = 2: 1cos2x-2: root sign 3sin2x + sin2x = 2: 1cos2x-2: root sign 3sin2x + 2: 1-cos2x = 2: 1-root sign 3sin2x

How to simplify y = cos (SiNx)?

The graph of this composite function is very difficult to draw. Even if it is drawn, it is not very useful to solve the problem. Does the building owner solve the problem of calculating the range of function? Don't take it out of context. It doesn't have to simplify the function to the way of understanding

Y = cos2x SiNx is reduced to 1-sin2x-sinx, and how to do it next is required to explain and solve the problem in detail, so that we can understand the 100% adoption!

Y = cos2x SiNx = 1-2sin & # 178; x-sinx = - 2 (Sin & # 178; + 1 / 2sinx + 1 / 16) + 1 / 8 + 1 = - 2 (SiNx + 1 / 4) & # 178; + 9 / 8 because - 3 / 4 ≤ SiNx + 1 / 4 ≤ 5 / 4, so 0 ≤ (SiNx + 1 / 4) & # 178; ≤ 25 / 16, so - 25 / 8 ≤ - 2 (SiNx + 1 / 4) & # 178; ≤ 0, so - 2 ≤ - 2 (SiNx + 1 / 4) & # 178; + 9

Given the function f (x) = sin (x + π) / 2, G (x) = Tan (π - x), which of the two functions is odd and which is even?
How do I have to make both functions odd

If f (x) = f (- x), then it is an even function
If - f (x) = f (- x), then it is an odd function
As long as it's based on this judgment
f(x)=sin(x+π)/2
f(-x)=sin(-x+π)/2=-sin(x-π)/2==-sin(x-π+2π)/2=-sin(x+π)/2=f(x)
So it's an even function
g(x)=tan(π-x)=-tanx
g(-x)=tan(π+x)=tanx=-g(x)
So it's an odd function

If f (x) = SiNx cosx and f ′ (x) = 2F (x), f ′ (x) is the derivative of F (x), then sin2x = ()
A. 13B. -35C. 35D. -13

∵ function f (x) = SiNx cosx and f ′ (x) = 2F (x), ∵ cosx + SiNx = 2sinx-2cosx, ∵ SiNx = 3cosx, ∵ TaNx = 3, ∵ sin2x = 2sinxcosx = 2sinxcosxsin2x + cos2x = 2tanxitan2x + 1 = 35