The vertex of the known angle α coincides with the origin of the rectangular coordinate system, and the starting edge is on the non negative half axis of the X axis It is known that the vertex of the angle a coincides with the origin of the plane rectangular coordinate system, the starting edge is on the non negative half axis of the X axis, and the final edge passes through the point P (- 1,2)

The vertex of the known angle α coincides with the origin of the rectangular coordinate system, and the starting edge is on the non negative half axis of the X axis It is known that the vertex of the angle a coincides with the origin of the plane rectangular coordinate system, the starting edge is on the non negative half axis of the X axis, and the final edge passes through the point P (- 1,2)

The solution consists of the vertex of the angle A and the origin of the plane rectangular coordinate system, the starting edge on the non negative half axis of the X axis, and the final edge passing through the point P (- 1,2)
Then Sina = 2 / √ 5, cosa = - 1 / √ 5
So sin (2a + 2 π / 3)
=sin2acos2π/3+cos2asin2π/3
=2sinacosa×（-cosπ/3）+（2cos^2a-1)×sinπ/3
=2×2/√5×（-1/√5）×（-1/2）+（2（-1/√5）^2-1)×√3/2
=2×2/√5×（1/√5）×（1/2）+（2×1/5-1)×√3/2
=2/√5×（1/√5）+（-3/5)×√3/2
=2/5-3√3/10
=（4-3√3）/10
Take the origin of rectangular coordinate system as the pole, and the positive half axis of X axis as the pole
Taking the origin of the rectangular coordinate system as the pole and the positive half axis of the x-axis as the polar axis, the same length unit is obtained in the two coordinate systems. The known polar coordinate equation of the straight line is η = π / 4 (P ∈ R), which intersects the curve X = 1 + 2cos α y = 2 + 2sina (α as the parameter) at points a and B, then | ab | is equal to
Sorry to be late, x = 1 + 2cos α y = 2 + 2sina cosa = (x-1) / 2sina = (Y-2) / 2 get (x-1) &# 178; + (Y-2) &# 178; = 4 the equation of the center of the circle is (1,2) the equation of the straight line is y = x the distance from the center of the circle to the straight line is | 1-2 | / √ 2 = √ 2 / 2 the radius is 2, so the half chord length is √ (4-1 / 2) = √ 14 / 2 the chord length is √ 14 | ab |
In the rectangular coordinate system, the polar coordinate system is established by taking the coordinate origin as the pole and the positive half axis of X axis as the polar axis. It is known that the polar coordinate equation of the straight line L is ρ = cos (θ - Π / 4) = √ 2, and the parameter equation of the curve C is x = 2cos θ y = sin θ (θ is logarithm)
ρ cos (θ - Π / 4) = √ 2 ρ cos θ cos Π / 4 + ρ sin θ sin Π / 4 = √ 2 (√ 2) / 2x + (√ 2) / 2Y = √ 2Y = - x + 2, i.e. the linear equation (x / 2) ^ 2 + y ^ 2 = 1 (x ^ 2) / 4 + y ^ 2 = 1 is substituted into the linear equation to get 5 / 4x ^ 2-4x + 3 = 0. Weida's theorem shows that the distance between two points of the abscissa (6 / 5,4 / 5) is not common except (2,0)
Taking the origin o of rectangular coordinate system as the pole and the positive half axis of X axis as the polar axis, the rectangular coordinates of known point P are (1, - 5) and the polar coordinates of point m are (1, - 5)
If the line L passes through point P and the inclination angle is, circle C takes m as the center and 4 as the radius
(1) The parametric equation of line L and the polar coordinate equation of circle C are solved;
(2) Try to determine the position relationship between line L and circle C
(1) ∵ the line L passes through the point P (1, - 5), and the inclination angle is π / 3 ∵ the parameter equation of the line L is x = 1 + 1 / 2T, y = - 5 + √ 3 / 2T (t is the parameter) ∵ the polar coordinate of the center of the circle C with radius 4 is (4, π / 2) ∵ the central coordinate is (0,4), and the rectangular coordinate equation of the circle is x ∧ 2 + (y-4) ∧ 2 = 16, so the polar coordinate of the circle C is
It is known that the function f (x) is a decreasing function in the domain (- infinity, 4) and can make f (m-sinx) ≤ f (under the root sign (1 + 2m) - 7 / 4 + + cos ^ 2x)
Find the range of M
Was the answer wrong
be like:
√(1+2m)-7/4+cos²x≤m-sinx.......(1)
√(1+2m)-7/4+cos²x≤4.........(2)
m-sinx≤4..........(3)
(1)===>√(1+2m)-m ≤sin²x-sinx+3/4=(sinx -1/2)²+1/2
It holds for any x ∈ R
===>√(1+2m)-m ≤1/2
m²+1/4+m≥1+2m
(m-1/2)²-1≥0
m≥3/2
Or m ≤ - 1 / 2
(2)==>√(1+2m)≤23/4-cos²x
For any x ∈ r = = > √ (1 + 2m) ≤ 19 / 4
-1/2≤m≤345/32
3) = = = = > m ≤ 4 + SiNx holds for any x ∈ r = = = = > m ≤ 3
The above intersection
==>m∈[3/2 ,3] or m=-1/2
Because f (x) is a decreasing function in the domain of definition (- ∞, 4), as long as
4≥√（1+2m） - 7/4 + Cos^2 X≥m-sinx
That's fine
(1)
From 4 ≥ √ (1 + 2m) - 7 / 4 + cos ^ 2 x →
1+2m≥0;→m≥-1/2;
If cos ^ 2 x ≤ 23 / 4 - √ (1 + 2m), then if cos ^ 2 x ≤ 1 and x = R, there must be
23/4-√（1+2m）≥1.
The solution is m ≤ 357 / 32
M ≥ - 1 / 2,
∴-1/2≤m≤357/32.
II.
4 ≥ m-sinx
sinx≥m-4;
If the range of trigonometric function SiNx ≤ 1 and x = R, then there must be
m-4≤1;
→m≤5.
3.
From √ (1 + 2m) - 7 / 4 + cos ^ 2 x ≥ m-sinx, it is concluded that: 1
Cos^2 X +sinx ≥m-√（1+2m）+ 7/4
Then - 2Sin ^ 2x + SiNx + 1 ≥ m - √ (1 + 2m) + 7 / 4
-2(sinx -1/4)^2 +9/8 ≥ m-√（1+2m）+ 7/4.
Then from the range of trigonometric function - 1 ≤ SiNx ≤ 1, we can get - 2 ≤ - 2 (SiNx - 1 / 4) ^ 2 + 9 / 8 ≤ 9 / 8
And if x = R, then there must be
-2≤ m-√（1+2m）+ 7/4≤9/8
These are two inequalities; solve them separately
From - 2 ≤ m - √ (1 + 2m) + 7 / 4, it is obtained that: 1
Tick (1 + 2m) ≤ m + 2 → square
1+2m≤ m^2+4m+4;
Then m ^ 2 + 2m + 3 ≥ 0; m ∈ R;
From M - √ (1 + 2m) + 7 / 4 ≤ 9 / 8, it is concluded that: 1
M-5 / 8 ≤ √ (1 + 2m); → square
1+2m≤ m^2-(5/4)m+25/64;
Thus M can be determined
Taking the intersection of ①, ② and ③ is the value range of real number M
//Yes, I calculate it according to the formula under the root sign, which is only (1 + 2m)
Ask your math teacher. I can't understand it....
Given A1 = 1,2an + 1 * an + 3an + 1 + an + 2 = 0, prove that {1 / an + 1} is the arithmetic sequence (2) to find an
There is a big ambiguity in the absence of brackets
(1)
2a(n+1)*a(n) + 3a(n+1) + a(n) + 2 = 0
2a(n+1)*a(n) + 2a(n+1) + 2a(n) + 2 = a(n) - a(n+1)
2[a(n+1) +1]*[a(n) + 1] = [a(n) + 1] - [a(n+1) + 1]
2 = 1/[a(n) + 1] - 1/[a(n+1) + 1]
Obviously {1 / [a (n) + 1]} is an arithmetic sequence with the first term of 1 / 2 and the tolerance of - 2
(2) From (1), a (n) = - 2n - 0.5 can be obtained
-2n - 0.5
The square of x minus 15 equals minus 56
Wrong. Wrong. It's 15x
That's right. The problem is wrong. The square of x minus 15x equals minus 56
The results are as follows
X^2-15X=-56
X^2-15X+56=0
(x-7)(x-8)=0
So x = 7 or x = 8
Is it possible
x^2-15=-56
x^2=-39
x=√39i
x^2=-41
X = positive and negative root sign 41I
According to the meaning of the title:
X*X-15=-56
The result is: X * x = - 56 + 15
Combining the similar items, we get x * x = - 41
……
How is that possible? Is your question wrong?!
The square of any number is positive. If you subtract 15 from a positive number, you can't get - 56.
This question is wrong.
First floor yeah, it's a plural question... Never been to high school?
The square of x minus 15x equals minus 56
=x^2-15x+56=0
(x-7)(x-8)=0
So x = 7 or 8
It is known that a = {x | x ＜ - 1 or X ＞ 5}. B = {x | a ≤ x ＜ a + 4}. If B is a proper subset of a, then the value range of real number a is?
If you don't understand this question, my question is: B is the proper subset of a, doesn't a contain B, and it's not equal to B? Isn't the number of a set - 2 - 3 - 4 - 5 or 67 8 9 10 11... Because B is contained in a, what I want to ask is: the beginning of B set is the same as a set - 2, If the lower row is a set: - 2-3-4-5-6, b set - 2-3-4-5, then a does not contain B? The other is that the beginning of B set is different from a set, and the beginning of B set is a number smaller than - 2. For example, a set - 2-3-4-5-6b set - 3-4-5-6, then a set does not contain b set, and the tail number of B set, that is - 6, can accompany a set to cycle until it is unclear, But the first method can't. in the first method, set B can only rank the number in front of mantissa with set a, such as set a: - 2-3-4-5-6, set b-2-3-4-5, which kind of bag is the cause of the answer? It's not the same puzzle. It seems that I'm in a corner. I really want to know the answer
First of all, the elements of set a and B are not all integers, and there are other numbers (such as irrational numbers),
Secondly, the elements in set B are only a small segment (although a can take all real numbers, once a is determined, the length of set B is only 4). Therefore, a and B cannot be equal
Because B is the proper subset of A,
So a + 45,
The solution is A5
In the range of monotone definition, f (a) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (?) (
Because the function domain is (- ∞, 3) and monotonically decreasing, 3 > A ^ 2-sin (x) > = a + 1 + cos (x) ^ 2 holds for all x, 3 > A ^ 2-sin (x) holds for all x, and the solution - √ 2 = 0 holds for all X. let y = sin (x), that is, the inequality y ^ 2-y + A ^ 2-a-2 > = 0 holds on [- 1,1]
If the sequence {an} satisfies a (n + 1) = 3an + n (n belongs to positive integer), is there A1 to make {an} an arithmetic sequence
Let {an} satisfy a (n + 1) = 3an + N, then A2 = 3A1 + 1; A3 = 3a2 + 2 = 3 (a1 + D) + 2. Therefore, a1 + 3 (a1 + D) + 2 = 2 (3A1 + 1) is reduced to A1 = 3D / 2. Since the tolerance is D, A2 = a1 + D = 5D / 2; A3 = A2 + D = 7d / 2; A3 = A2
a(n+1) = 3a(n) + n, n = 1,2,...
set up
a(n+1) + b(n+1) + c = 3[a(n) + bn + c],
be
n = 3bn + 3c - bn - b - c = 2bn + 2c - b,
b = 1/2, c = b/2 = 1/4.
A (n + 1) + (n + 1) / 2 + 1 / 4 = 3 [a (n) +... Expansion
a(n+1) = 3a(n) + n, n = 1,2,...
set up
a(n+1) + b(n+1) + c = 3[a(n) + bn + c],
be
n = 3bn + 3c - bn - b - c = 2bn + 2c - b,
b = 1/2, c = b/2 = 1/4.
a(n+1) + (n+1)/2 + 1/4 = 3[a(n) + n/2 + 1/4],
Let B (n) = a (n) + n / 2 + 1 / 4, n = 1,2
Then B (n) = 1
{B (n)} is an equal ratio sequence whose first term is B (1) = a (1) + 1 / 2 + 1 / 4 = a (1) + 3 / 4, and the common ratio is 3.
b(n) = [a(1)+3/4]3^(n-1), n = 1,2,...
a(n) = b(n) - n/2 - 1/4 = [a(1)+3/4]3^(n-1) - n/2 - 1/4, n = 1,2,...
a(n+1) = [a(1)+3/4]3^n - (n+1)/2 - 1/4
a(n+1) - a(n) = [a(1)+3/4]3^n - (n+1)/2 - 1/4 - {[a(1)+3/4]3^(n-1) - n/2 - 1/4}
= 2[a(1)+3/4]3^(n-1) - 1/2
In order to make a (n + 1) - A (n) independent of N, only a (1) = - 3 / 4
When a (1) = - 3 / 4,
a(n) = -n/2 - 1/4. n = 1,2,...
The number sequence is equal to (a)}. Put it away