The range of F (x) = (1 + Sina) (1 + COSA) (a is a real number)

The range of F (x) = (1 + Sina) (1 + COSA) (a is a real number)

F (x) = (1 + Sina) (1 + COSA) = 1 + sinacosa + Sina + cosa this is one of the methods: let t = SiNx + cosx then - root 2 ≤ t ≤ square of root 2 get, T ^ 2 = 1 + 2sinxcosx, sinxcosx = 1 / 2 (t ^ 2-1) can be substituted to get f (x) = 1 + (T ^ 2-1) / 2 + tmaximum value is 3 / 2 - root 2
The parametric equations are transformed into ordinary equations x = 1 + cosa, y = Sina + Tana
X=1+cosA cosA=x-1
y=sinA+tanA=sinA(1+1/cosA)=sinA(x/(x-1)) sinA=y(x-1)/x
Square addition
(x-1)^2+y^2(x-1)^2/x^2=1
y/x=(sina+tana)/(1+cosa)=sina
So y & # 178 / / X & # 178; = Sin & # 178; a
cosa=x-1
cos²a=x²-2x+1
So y & # 178; / X & # 178; + X & # 178; - 2x + 1 = 1
y²=-x^4+2x³
X=1+cosA，y=sinA+tanA=tanA*(cosA+1)
So we have y = x * Tana, that is, Tana = Y / X
And 1 + Tan ^ 2A = 1 / cos ^ 2A
Therefore: 1 + y ^ 2 / x ^ 2 = 1 / (x-1) ^ 2
The common equation is: 1 / (x-1) ^ 2-y ^ 2 / x ^ 2 = 1
How can Sina cosa / Sina + cosa become Tana / Tana + 1
To be specific steps
The numerator and denominator are divided by cos ^ 2A at the same time
=(sinacosa/cos^2a)/(sin^2a/cos^2a+cos^2a/cos^2a)
=tana/(tan^2a+1)
In polar coordinate system, the distance from point (1,0) to line P (COSA + Sina) = 2 is______________ (hope to analyze)
Simplify P (COSA + Sina) = 2 to x + Y-2 = 0, and then use the distance formula from point to line ~!
Note that the product of the first n terms of sequence an is TN = 1-an, and that CN = 1 / TN. the sum of the first n terms of sequence BN is Sn and Sn = 1-bn. (1) prove that CN is an arithmetic sequence; (2) if TN (NbN + n-2) ≤ kn is a positive integer constant for N, find the value range of real number K
(1) When n = 1, T1 = 1-a1 = A1, A1 = 1 / 2
When n = 2, T2 = A1 * A2 = (1 / 2) A2 = 1-a2, A2 = 2 / 3
When n = 3, T3 = A1 * A2 * A3 = (1 / 3) A3 = 1-a3 A3 A3 = 3 / 4
Suppose that AK = K / (K + 1) TK = 1-ak holds when n = K,
When n = K + 1, t (K + 1) = TK * a (K + 1) = (1-ak) * a (K + 1) = [1 / (K + 1)] * a (K + 1) = 1-A (K + 1)
a(k+1)=(k+1)/(k+2)
So an = n / (n + 1) TN = 1-an = 1 / (n + 1) CN = n + 1
(2) Similarly, BN = (1 / 2) ^ n can be obtained
[1/(n+1)][n(1/2)^n+n-2]≤kn n(1/2)^n+n-2≤kn^2+kn n(1/2)^n≤kn^2+(k-1)n+2
(n + 1) (1 / 2) ^ (n + 1) - n (1 / 2) ^ n = (1-N) (1 / 2) ^ (n + 1) ≤ 0 and equal when n = 1, and subtracted when n > 1
So the maximum value of n (1 / 2) ^ n is 1 / 2 when n = 1 or 2
Let f (n) = kn ^ 2 + (k-1) n + 3 / 2 ≥ 0
① When k = 0, there is - N + 3 / 2 ≥ 0, only when n = 1
② K > 0 axis of symmetry - (k-1) / 2K ≤ 1 K ≥ 1 / 3 F (1) = 2K + 1 / 2 ≥ 0 K ≥ - 1 / 4
So K ≥ 1 / 3
In the triangle ABC, a = 60 & # 186;, angle B ＞ angle c, B, C are the two real roots of the equation x & # 178; - 2 √ 3 + m, the area of triangle ABC is √ 3 / 2, and the length of three sides is obtained
According to B, C is the solution of the equation x & # 178; - 2 √ 3 + M = 0, we can get: B + C = 2 √ 3, BC = m; while the area of triangle ABC is √ 3 / 2, then s △ = 1 / 2 (bcsina) = √ 3 / 2, then 1 / 2 (m * √ 3 / 2) = √ 3 / 2, then M = 2; and because ∠ B ＞ C, then B + C = 2 √ 3, BC = m, then B = √ 3 + 1, C = √ 3-1
Given that the equation x & # 178; - 2aX + A & # 178; - 1 = 0 has roots in the interval [- 2,4], then the value range of the real number a
x²-2ax+a²-1=0
(x-a+1)(x-a-1)=0
x1=a-1 x2=a+1
According to the meaning of the title: - 2 ≤ A-1 ≤ 4 ① Or - 2 ≤ a + 1 ≤ 4 ② Or A-1 ≥ - 2 and a + 1 ≤ 4 3.
From ①: - 1 ≤ a ≤ 5; from ②: - 3 ≤ a ≤ 3; from ③: - 1 ≤ a ≤ 3
The union of the three results in: - 3 ≤ a ≤ 5
Given that the function f (x) = LG (2 / 1-x a) is an odd function, find the inequality f (x)
f(x)=lg(2/1-x+a)=lg[(2+a-ax)/(1-x)]
f(-x)=lg(2/1-x+a)=lg[(2+a+ax)/(1+x)]=-f(x)
[(2+a-ax)/(1-x)]＊[(2+a+ax)/(1+x)]=1
a=-1
f(x)=lg(1+x)/(1-x)
Sn = n ^ 2, let BN = an / 3 /, and let the sum of the first n terms of the sequence {BN} be TN
Given the sum of the first n terms of the sequence an and Sn = n ^ 2, let BN = an / 3 ^ n, note that the sum of the first n terms of the sequence BN is TN
Verification: TN = 1 - (n + 1) / 3 ^ n
Although it's copy and paste
But I still don't understand.
A (1) = s (1) = 1, n > 1, a (n) = s (n) - S (n-1) = n ^ 2 - (n-1) ^ 2 = 2N-1, a (n) = 2N-1, n = 1,2,... B (n) = a (n) / 3 ^ n = (2n-1) / 3 ^ n, n = 1,2,... [the sum of B (n) is divided into two parts, Part 1 is the sum of 2n / 3 ^ n, Part 1 is the sum of 1 / 3 ^ n.The sum of Part 2 is very simple
Then you go to see something called dislocation subtraction
It is known that a, B and C are the three sides of △ ABC, and two of the equations a (1 + x2) + 2bx-c (1-x2) = 0 are equal, then the shape of △ ABC is:______ .
The original equation is (a + C) x2 + 2bx + a-c = 0, because the two are equal, so △ = b2-4ac = (2b) 2-4 × (a + C) × (A-C) = 4B2 + 4c2-4a2 = 0, that is, B2 + C2 = A2, so △ ABC is a right triangle