Given that π / 2 < a < π, cos (π / 4-A) cos (π / 4 + a) = 1 / 8, find the value of Tana

Given that π / 2 < a < π, cos (π / 4-A) cos (π / 4 + a) = 1 / 8, find the value of Tana

I remember answering this question. Why did it appear in the team help again? Copy my answer as follows: cos (π / 4-A) cos (π / 4 + a) = cos (π / 4-A) sin (π / 4-A) = sin (π / 2-2a) / 2 = (cos2a) / 2 = 1 / 8, so cos2a = 1 / 4sin & # 178; 2A = 1-cos & # 178; 2A = 1-1 / 16 = 15 / 16 π / 2 < a < π
It is known that cos (3, pai-a) = - 4 / 5, a ∈ (0, Pai,) then Tana=
3/4
Known 0
Zero
Zero
Given that the minimum positive period of the function y = sin (Wx + π / 4) is 2 π / 3, then w is equal to ()
Given that the minimum positive period of the function y = sin (Wx + π / 4) is 2 π / 3, then w is equal to (3)
Because: the minimum positive period of y = sin (Wx + π / 4) is 2 π / 3
So: 2 π / w = 2 π / 3
The solution is: w = 3
T=2π/W=2π/3 W=3
The real number ABCD is consistent with a plus B equals C plus D, the cube of a plus B equals the cube of C plus D, and the 2010 power of a plus B equals the 2010 power of c2010 plus D
a+b=c+d
a^3+b^3=c^3+d^3
(a+b)(a^2+b^2-ab)=(c+d)(c^2+d^2-cd)
a^2+b^2-ab=c^2+d^2-cd
(a+b)^2-3ab=(c+d)^2-3cd
ab=cd
a. B, C and D are the same equations
x^2-(a+b)x+ab=0
The root of the tree
If a is equal to one of C and D, then B must be equal to the other of C and D. let a = C and B = D
There are: A ^ 2010 + B ^ 2010 = C ^ 2010 + D ^ 2010
The proof is complete
In fact, the Index 2010 to any real number, the equation is true
Given that the distance between the nearest two intersections of the intersection of the image of the function f (x) = 2Sin (Wx + π / 5) and the line y = - 1 is π / 3, then the value of W?
2sin(ωx+π/5) =-1
sin(ωx+π/5)=-1/2
(ω x + π / 5) = 2K π - π / 6 or 2K π - π 5 / 6
X = (2k π - 11 π / 30) / ω or (2k π - 31 π / 30) / ω
So: π / 3 = (2k π - 11 π / 30) / ω - (2k π - 31 π / 30) / ω
That is: π / 3 = 2 π / 3 ω
So: ω = 2
When π = 5 π / sin = 5 π + and π = 5 π / sin = 5 π / sin
So: Wx + π / 5 = 7 π / 6 and w (x + π / 3) + π / 5 = 11 π / 6
The results show that w * (π / 3) = 11 π / 6-7 π / 6 = 2 π / 3
So: w = 2
Yeah. - right
2sin(ωx-π/5) =-1
sin(ωx-π/5)=-1/2
(ωx-π/5)=2Kπ-π/6，2Kπ-π5/6
ω(π/3)=2π/3
ω=2
Given that the cube of a + the cube of B = 27, the square of a times B-A times (the square of B) = - 6
Find the value of (B's cube-a's Cube) + (A's Square times B-3A times B's Square.) - 2 times (B's cube-a's Square times b)
Because B \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\- 33
F
G
-45
Given that the distance between the nearest two points in the intersection of the image of the function f (x) = 2Sin (ω x + φ) (ω > 0) and the line y = 1 is π 3, then ω is equal to______ .
Let ω X1 + φ = π 6 + 2K π, K ∈ Z & nbsp; ① ω x2 + φ = 5 π 6 + 2K π, (K ∈ z) ②, known as: x2-x1 = π 3. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ② - ①, we get: ω = 2. So the answer is: 2
How much is the square of a times the cube of - a?
-The fifth power of a
-The fifth power of a
The same base power subtraction, exponential division
Two thirds power of a
-The fifth power of a
-The fifth power of a
Given the function f (x) = 2Sin (Wx + V), the distance between the nearest two points in the intersection of the image and the line y = 1 is 60 ° to find W
∵ the distance between the nearest two points in the intersection of y = 1 and the line is π / 3
∴f(kπ+π/2)-f(kπ+π/2-π/6)=1
That is, 2Sin [ω (K π + π / 2) + φ] - 2Sin [ω (K π + π / 2 - π / 6) + φ] = 1
And f (K π + π / 2) = 2
f(kπ+π/2-π/6)=1
The solution is ω = 1