Let a = (1,1) B = (cosx, SiNx) find the maximum value and period of the function f (x) = a · B. If a · B = 1 / 2, find the value of (2Sin ^ 2x + sin2x) / (1 + TaNx)

Let a = (1,1) B = (cosx, SiNx) find the maximum value and period of the function f (x) = a · B. If a · B = 1 / 2, find the value of (2Sin ^ 2x + sin2x) / (1 + TaNx)

(1) If f (x) = a · B = cosx + SiNx = Radix 2Sin (x + 45), the maximum value is root 2 and the period is 2 π
(2) F (x) = a · B = cosx + SiNx = 1 / 2 sin? X + cos? X = 1, so 2cosxsinx = - 3 / 4
（2sin^2x+sin2x)/（1+tanx）
=（2sin²x+2cosxsinx）/（1+sinx/cosx）
=(2Sin? X-3 / 4) / (1 + SiNx / cosx) multiply cosx up and down
=(2Sin? Xcosx-3 / 4cosx) / 1 / 2
=（sinx*-3/4-3/4cosx）/1/2
=-3/4

The function f (x) = sin2x+ The minimum positive period of 3cos2x is______ .

It can be concluded from the meaning of the title
y=sin2x+
3cos2x
=2（ 1
2sin2x+
Three
2cos2x）
=2sin（2x+π
3）
∴T=2π
2=π
So the answer is: π

The minimum positive period and maximum value of the function f (x) = (sin2x-cos2x) 2 are

f(x)=(sin2x-cos2x)^2
=sin²2x+cos²2x-2sin2xcos2x
=1-sin4x
So the minimum positive period T = 2 π / 4 = π / 2,
When sin4x = - 1, the original function has a maximum value of 2

The minimum positive period and maximum value of the function f (x) = (sin2x-cos2x) 2 are Please be detailed, as well as the formula, thank you

f(x)=(sin2x-cos2x)²
=sin²2x+cos²2x-2sin2xcos2x
=1-sin4x
Minimum positive period = 2 π / 4 = π / 2
Max = 1 + 1 = 2

The minimum positive period and maximum of the quadratic power of the function f {x} = {sin2x cos2x} Please explain the solution and process

f｛x｝=（sin2x-cos2x）²
= 1 - 2sin2xcos2x
= 1 - sin4x
So the minimum positive period is 2 π / 4 = π / 2
The maximum value is 2 when sin4x = - 1

Given that the function f (x) = sin ^ x + 2sinxcosx + 3cos ^ x, X belongs to R, find 1: the maximum value of function f (x) and the set of independent variables X when obtaining the maximum value; 2: calculate The monotone increasing interval of function f (x)

One
f(x)=(sinx)^2+2sinxcosx+3(cosx)^2
=1+sin2x+2(cosx)^2
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
So the maximum value of F (x) is 2 + √ 2
When 2x + π / 4 = 2K π + π / 2 (K ∈ z), that is, x = k π + π / 8 (K ∈ z), the maximum value of F (x) is taken
Therefore, the set of independent variables X with maximum value is {x | x = k π + π / 8 (K ∈ z)}
Two
Let 2K π - π / 2 < 2x + π / 4 < 2K π + π / 2 (K ∈ z)
K π - 3 π / 8 < x < K π + π / 8 (K ∈ z) is obtained
So the monotone increasing interval of function f (x) is (K π - 3 π / 8, K π + π / 8) (K ∈ z)
If you do not understand, please hi me, I wish you a happy study!

Find the maximum value, minimum value and period of the function y = sin2x + √ 3 * cos2x, and find the set of X that makes the function get the maximum and minimum value

y=2(sin2xcosπ/3+cos2xsinπ/3)
=2sin(2x+π/3)
therefore
Max = 2, min = - 2
Period = 2 π / 2 = π
At the maximum value, 2x + π / 3 = 2K π + π / 2
2x=2kπ+π/6
x=kπ+π/12
That is, the set is {x | x = k π + π / 12, K ∈ Z}
At the minimum value, 2x + π / 3 = 2K π + 3 π / 2
2x=2kπ+7π/6
x=kπ+7π/12
That is, the set is {x | x = k π + 7 π / 12, K ∈ Z}

Given the function f (x) = 2 + sin2x + cos2x, find the monotone increasing interval of function f (x)

First, the original function is converted into a sine function f (x) = sin2x + cos2x + 2 = (sqrt2) * (sin2x * (sqrt2) / 2 + cos2x * (sqrt2) / 2 + cos2x * (sqrt2) / 2) + 2 = (sqrt2) * sin (2x + π / 4) + 2, when ω x + φ belongs to [2K π - π / 2, 2K π + π / 2], the function monotone increases, that is, 2x + π / 4 belongs to [2K π / 2, 2K π + π / 2], that is, 2x2x + π / 4 belongs to [2K π / 2, 2K π + π / 2], that is, 2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x2x x belongs to 2K π - 3 π / 4, 2K π + π / 4 increases monotonically, K π + π / 8] if you have any questions about sine function, you can ask it. Hope it can help you!

Let the minimum value of the function y = 2cos ^ 2x-2acosx - (2a + 1) be f (a), and find the expression of F (a) I can't understand this step When - 1 ≤ A / 2 ≤ 1, that is - 2 ≤ a ≤ 2, yman = - (a ^ 2 / 2 + 2A + 1) When a / 2 > 1, that is, a > 2, Ymin = y|x = 1 = 2-2a - (2a + 1) = - 4A + 1 When a / 2 There are also sections by what

Let x = cosx, so y = 2x ^ 2-2ax - (2a + 1). According to the symmetry axis of parabola x = - B / 2a, x = A / 2
When x = cosx, the value of X is limited to [- 1,1], and you should understand

Given that x ∈ [0, π / 2], (1) find the minimum value of the function y = (cosx) ^ 2-2acosx; (2) find the function y = 3sinxcosx-3 √ 3 (SiNx) ∧ 2 + (3 √ 3)/ Given x ∈ [0, π / 2], (1) find the minimum value of the function y = (cosx) ^ 2-2acosx (2) Find the maximum and minimum of the function y = 3sinxcosx-3 √ 3 (SiNx) ∧ 2 + (3 √ 3) / 2

1,y=(cosx)^2-2acosx=(cosx-a)^2-a^2
Zero