Cosa = - cos ^ 2 A / 2, then what is cosa / 2 equal to

Cosa = - cos ^ 2 A / 2, then what is cosa / 2 equal to

cosa=-cos²(a/2)
=2cos²(a/2)-1
therefore
3cos²(a/2)=1
Cos (A / 2) = ± 3 / 3
Given cosa = 1 △ 3, cos (a + b) = - 1 △ 3, and a, B ∈ (0, π △ 2), then the value of COS (a-b) is equal to
In the evening,
∵a,b∈(0,π÷2)
∴sina>0 sin(a+b)>0
∴sina=√(1-cos²a)=√[1-(1/3)²]=2√2/3
cos(a+b)=1/3cosb-2√2/3sinb=-1/3 ①
∵sin(a+b)=√[1-(-1/3)²]=2√2/3
∴2√2/3cosb+1/3sinb=2√2/3 ②
From the simultaneous solution of (1) and (2)
cosB=7/9,sinB=4√2/9
∴cos(A-B)=1/3cosB+2√2/3sinB=23/27
It is known that cosa = 1 ／ 3, cos (a + b) = - 1 ／ 3, and a, B ∈ (0, π ／ 2)
Then a + B ∈ (0, π)
Sina = radical (1-cos & # 178; a) = 2 (radical 2) / 3
Sin (a + b) = radical [1-cos & # 178; (a + b)] = 2 (radical 2) / 3
So: sin2a = 2sina * cosa = 2 * 2 (radical 2) / 3 * 1 / 3 = 4 (radical 2) / 9
Cos2a = 1-2cos... Expansion
It is known that cosa = 1 ／ 3, cos (a + b) = - 1 ／ 3, and a, B ∈ (0, π ／ 2)
Then a + B ∈ (0, π)
Sina = radical (1-cos & # 178; a) = 2 (radical 2) / 3
Sin (a + b) = radical [1-cos & # 178; (a + b)] = 2 (radical 2) / 3
So: sin2a = 2sina * cosa = 2 * 2 (radical 2) / 3 * 1 / 3 = 4 (radical 2) / 9
cos2a=1- 2cos²a=1- 2/9=7/9
So:
cos(a-b)=cos[2a-(a+b)]
=cos2a*cos(a+b) +sin2a*sin(a+b)
=(7 / 9) * (- 1 / 3) + [4 (radical 2) / 9] * [2 (radical 2) / 3]
=-7/27 + 16/27
=9/27
=1 / 3 "put away
The following statement is correct: A. The equation x ^ 2 + 1 = 0 has no root; B. pure imaginary numbers and imaginary numbers form the set of real numbers; C. the set of real numbers has imaginary numbers and complex numbers form the set of real numbers
The following statement is true
A. The equation x ^ 2 + 1 = 0 has no roots
B. Pure imaginary numbers and imaginary numbers constitute the set of real numbers
C. The set of real numbers consists of imaginary numbers and complex numbers
D. Real numbers are plural numbers
A. There are imaginary roots B, real numbers are composed of integers and fractions C, real numbers and imaginary numbers are composed of complex numbers, so D is chosen
D
D
The equation of straight line and circle, elliptic hyperbolic parabola in the college entrance examination score!
What is the proportion of these two parts in the college entrance examination?
It seems to be the two most difficult parts of high school mathematics!
seek
It should be said that the most difficult question in the college entrance examination is the last comprehensive question. We often mix functions and sequences together, and analytic geometry generally appears as the penultimate or the third question. Of course, it's not too difficult. Calculate 12 points. In addition, there will be about 2 analytic geometries, 10 points. If you fill in the blank, there may be one, 4 points. In this way, there will be 30 points
In the range of real numbers, the solution of the equation x & # 178; + 5 = 0 is a set without any elements, which is an empty set & # 8709;. Why?
Solution X & # 178; = - 5, x = (- 5) square root x = a magic horse number? Irrational number or rational number?
x^2=-5
X ^ 2 = 5i ^ 2 (I ^ 2 = - 1, I is plural)
X = ± root 5I
Not in the range of real numbers
Complex numbers contain real numbers
real number
It's the plural,
So that's to limit you to real numbers. There is no solution!
It's in the plural
X = ± root 5I
Quasi senior one? It's definitely not irrational number and rational number, otherwise the contradiction between the former (in the range of real number, the equation x & # 178; + 5 = 0 has no solution) and the latter is imaginary number, which should be learned in senior two
It is known that the parabola y = x ^ 2-2 and the ellipse x ^ 2 / 4 + y ^ 2 = 1 have four vertices, which are in common circle
Do not use dead reckoning, I just want to know what simple algorithm, thank you!
If we say a simple algorithm, we can only say that the circle is symmetric about the Y axis, so we can shoot the circle equation as
(Y-A) * (Y-A) + X * x = b * B, the rest is to use the traditional method to solve the intersection coordinates, and then solve the problem
A and B
Let m (x ^ 2-y-2) + (x ^ 2 / 4 + y ^ 2-1 = 0 L, let m + 1 / 4 = 1, then M = 3 / 4
The equation of the circle is x ^ 2 + y ^ 2-0.75y-2.5 = 0
Is imaginary number the same as real number?
It's about the same. It's a positive sign when even I are multiplied
An odd number of I is a minus sign when multiplied
Imaginary number and imaginary number operation, real number and real number operation, can't cross operation
The right focus of the ellipse is the same as that of the parabola y with square = 16, and the eccentricity is equal to 2 / 2 of the root sign
Right focus (4,0)
C=4
c/a=1/√2,a=4√2
b²=a²-c²=16
The equation of ellipse: X & # 178 / 32 + Y & # 178 / 16 = 1
What is the concept of imaginary number relative to real number?
In mathematics, if the square of a number is negative, that number is imaginary. All imaginary numbers are complex
The term "imaginary number" was coined by Descartes, a famous mathematician in the 17th century. At that time, it was thought that it was a real number that did not exist. Later, it was found that the imaginary number could correspond to the vertical axis of the plane, which was as real as the real number corresponding to the horizontal axis of the plane. The plane formed by the imaginary number axis and the real number axis was called the complex plane, and each point on the complex plane corresponds to a complex number
Let the hyperbola whose center is at the origin have a common focus with the hyperbola 2x ^ 2-2y ^ 2 = 1, and the sum of their eccentricities is 2 + radical 2, then the equation of the hyperbola is obtained
The known hyperbola is: x ^ 2 / (1 / 2) - y ^ 2 / (1 / 2) = 1, a = √ 2 / 2, B = √ 2 / 2, C = √ (a ^ 2 + n ^ 2) = 1, eccentricity E1 = C / a = 1 / (√ 2 / 2) = √ 2, another hyperbola's eccentricity E2 = 2 + √ 2 - √ 2 = 2, the two hyperbolas have common focus, then C = 1, E2 = C / A, 1 / a = 2, a = 1 / 2, B ^ 2 = C ^ 2-A ^ 2 = 1 / 4 = 3 / 4