Given Tana = 2, find the value of COS ^ 4 ((π / 3) + a) - cos ^ 4 ((a - π) / 6) Given Tana = 2, find the value of COS ^ 4 ((π / 3) + a) - cos ^ 4 (a - (π / 6)) The original question is wrong. It's this one

Given Tana = 2, find the value of COS ^ 4 ((π / 3) + a) - cos ^ 4 ((a - π) / 6) Given Tana = 2, find the value of COS ^ 4 ((π / 3) + a) - cos ^ 4 (a - (π / 6)) The original question is wrong. It's this one

cos^4((π/3)+a)-cos^4（a-(π/6)）=cos^4((π/3)+a)-cos^4（(π/6)-a）=cos^4((π/3)+a)-sin^4（(π/3)+a）=[cos^2((π/3)+a)+sin^2（(π/3)+a)]*[cos^2((π/3)+a)-sin^2（(π/3)+a)]=cos^2((π/3)+a)-sin^2（(π/3...
Find the known value of (Π) = (Π / A, a - ∈ 2)
sinA/cosA=-√2 sinA=-√2cosA
sin^2A+cos^2A=1
3cos^2A=1 A∈(∏/2,∏),cosA
The known function f (x) = sin (Wx + a) (a > 0, - π / 2)
(1) The distance between two adjacent highest and lowest points of function f (x) is 2 √ 2
According to the Pythagorean theorem, the difference between the ordinates of two points is known
The abscissa distance between two points is 2
The period of the function is 4
SO 2 π / w = 4 gives w = π / 2
And ∵ function f (x) passes (2, - 1 / 2) points
∴-1/2=sin(π+a)
∵-π/2≤a≤π/2
∴a=π/6
The analytic expression of function f (x) is f (x) = sin (π / 2 * x + π / 6)
(2) 2kπ-π/2≤π/2*x+π/6≤π/2+2kπ k∈Z
The solution is 4k-4 / 3 ≤ x ≤ 4K + 2 / 3
The monotone increasing interval of function f (x) is [4k-4 / 3,4k + 2 / 3]
Given A-B = 1, find the value of a to the third power + 3ab-b to the third power,
1-6ab
Given the function f (x) = sin (2wx-30 °) - 4sin Λ & #178; Wx + a (W > 0), the distance between the two adjacent highest points of the image is π. 1. Find the monotone increasing interval of function f (x). 2. Let the minimum value of function f (x) on [0,90 °] be - 3 / 2, find the range of function f (x) (x belongs to R)
(1) [2K π - 5 π / 12,2k π + π / 12] K is an integer sin (2wx - π / 6) = sin2wx * cos π / 6 - cos2wx * sin π / 6 = √ 3 / 2 * sin2wx - 1 / 2 * cos2wx. Because cos2wx = 1-2sin ^ 2wx, 4sin ^ 2wx = 2 - 2cos2wx f (x) = √ 3 / 2 * sin2wx - 1 / 2 * cos2wx - (2 -...)
Let a = the square of 1 + 2 * x, B = the cube of 2 * x + the square of X, X be a real number, and find the size relation of A.B
The answer in the book is a > = B
A-B=1+2X^2-2X^3-X^2
=-2X^3+X^2+1
=-(X-1)(2X^2+X+1)
=(2X^2+X+1)(1-X)
Because 2x ^ 2 + X + 1 = 2 (x + 1 / 4) ^ 2 + 7 / 8 > 0
So 1-x > 0 is XB;
1-x = 0, that is, when x = 1, a = B;
1-x1, a
The image of the function y = sin (2x - π / 3) and the line y = m have innumerable intersections, and the distance between any two adjacent intersections is equal=
When m = 0, the problem is satisfied, because sin (2x - π / 3) = 0, 2x - π / 3 = k π, x = k π / 2 + π / 6, when k is an integer, a series of X values will be obtained, and the distance between each adjacent intersection is equal, and all are = π / 2
It is known that a and B belong to positive real numbers. It is proved that the cube of a plus the cube of B is equal to a square times B plus a square
Are you wrong, let a = 1, B = 2, cube a plus cube B equals 9, square a times square B plus square a equals 3, obviously 9 is not equal to 3, so I don't think it's right
Given the function f (x) = √ 3 * sin ω x + cos ω x (ω ＞ 0), the distance between the image of F (x) D and the two adjacent intersections of the line y = 2 is π
Monotone increasing interval of F (x)
f(x)=2sin(wx+π/6)
So the maximum value of the function is 2
So the period is two
We get w = 2
Finding monotone interval below
-π / 2 + K π 2x + π / 6 π / 2 + K π (k belongs to integer) can solve the inequality
a. B and C belong to real numbers, and a is greater than B. why is it wrong that the cube of a is greater than the cube of B?