What is cos (a - π / 2) equal to

What is cos (a - π / 2) equal to

cos(a-π/2)=
sina
What is cos = / θ = 1,
±60°
What degree of COS is equal to 1
0 + 2kpai (k to integer)
The analytic inverse function is x (H) = x (1, H) + G (1)
F (x) is a positive proportional function and G (x) is an inverse proportional function
So let f (x) = ax, G (x) = B / X
So h (x) = ax + B / X
H (1) = 8, H (1-3) = 16, so substituting (1,8) (- 2,16) is OK
It is known that the two focal points of the ellipse are F1 (- 1,0) and F2 (1,0), P is a point on the ellipse, and | F1F2 | is the median of the difference between | Pf1 | and | PF2 | (1) solve the elliptic equation; (2) if point P satisfies ∠ f1pf2 = 120 °, find the area of △ pf1f2
(1) Let the elliptic equation be x2a2 + y2b2 = 1 (a > 0, b > 0). From the known | F1F2 | = 2, | Pf1 | + | PF2 | = 4 = 2A, | a = 2, B2 = a2-c2 = 4-1 = 3. The elliptic equation is x24 + Y23 = 1 (2). In △ pf1f2, | Pf1 | = m, | PF2 | = n. from the cosine theorem, we get 4 = M2 + n2-2mncos 120 degree
The definition field of X + 1 / 1 of X-1 of function y = square root 5 is
"The root of 5 is divided into X-1 and X-1"?
Analytical formula do not understand, denominator is not equal to 0, the number of root is greater than 0, intersection
Given the function H (x) = f (x) + G (x), where f (x) is the positive proportion function of X, G (x) is the inverse proportion function of X, and H (1 / 3) = 16, H (1) = 18, find H (x)
To find the analytic expression of H (x)
Let H (x) = ax + B / X
Then H (1 / 3) = A / 3 + 3B = 16
h(1)=a+b=18
a=57/4 b=15/4
h(x)=57/4*x+15/(4x)
Let the left and right focus of ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) be F1 and F2 respectively, and point P (a, b) satisfy PF2 = F1F2
(2) Let the line PF2 and the ellipse intersect at two points a and B. if the line PF2 and the circle (x + 1) ^ 2 + (y-radical 3) ^ 2 = 16 intersect at two points m and N, and Mn = (5 / 8) * AB, the elliptic equation is solved
(1) Let F 1 (- C, 0), F 2 (C, 0) (c > 0)
From the title, we can get | PF2 | = | F1F2 |, that is, the root sign (A-C) &# 178; + B & # 178; = 2C
∴2(c/a)²+c/a-1=0
C / a = 1 / 2 or - 1 (round)
∴e=1/2
(2) From (1) we know that a = 2c, B = root 3C, elliptic equation is 3x & # 178; + 4Y & # 178; = 12C & # 178;, linear equation PF2 is y = root 3 (x-C)
A. The coordinates of B satisfy the equations: 3x & # 178; + 4Y & # 178; = 12C & # 178;
Y = radical 3 (x-C)
The solution is x = 0, x = 8C / 5
The solution of the equations is: x = 0, y = - radical 3
X = 8C / 5, y = (3 radical 3 / 5) C
Let a (8C / 5, (3 radical 3 / 5) C), B (0, (- radical 3) C)
So | ab | = radical [(8C / 5) &# 178; + {[(3 radical 3 / 5) C + radical 3C] &# 178;}
So | Mn | = 8 / 5 | ab | = 2C
The distance from the center (- 1, radical 3) to the straight line PF2 d = | - radical 3-radical 3-radical 3C | / 2
∵d²+(|MN|/2)²=4²
∴3/4(2+c)²+c²=16
The solution is C = 2 or - 26 / 7
The elliptic equation is X & # 178 / 16 + Y & # 178 / 12 = 1
If the domain of the function y = (2 − a) x2 + 2 (2 − a) x + 4 is r, then the value range of a is r______ .
According to the meaning of the question: (2-A) x2 + 2 (2-A) x + 4 ≥ 0, X ∈ R is tenable. ① when 2-A = 0, 4 ≥ 0 is tenable. ② when 2-A > 0, △ = 4 (2-A) 2-16 (2-A) ≤ 0-2 ≤ a < 2. To sum up: - 2 ≤ a ≤ 2, so the answer is: - 2 ≤ a ≤ 2
Given that f (x) is a positive scale function and G (x) is an inverse scale function, and f (1) / g (1) = 2, f (2) + 4G (2) = 6, find
The analytic expressions of F (x) and G (x)
f(x)=2x
g(x)=1/x