Given the vector a = (2cosx / 2, Tan (x / 2 + π / 4)), B = (Radix bisin (x / 2 + π / 4)), Tan = (x / 2-π / 4) Let f (x) = a * B, Find the maximum value and minimum positive period of function f (x), and write the monotone interval of F (x) in [0, π]

Given the vector a = (2cosx / 2, Tan (x / 2 + π / 4)), B = (Radix bisin (x / 2 + π / 4)), Tan = (x / 2-π / 4) Let f (x) = a * B, Find the maximum value and minimum positive period of function f (x), and write the monotone interval of F (x) in [0, π]

a=(2cosX/2,tan(X/2+π/4))=（2cosx/2,（1+tanx/2）/（1-tanx/2））
B = (Radix Di sin (x / 2 + π / 4)), Tan = (x / 2-π / 4) = (SiNx / 2 + cosx * 2, (TaNx / 2-1) / (1 + TaNx / 2))
f(x)=a*b=sinx+cosx+1-1=√2sin（x+π/4）
So the maximum value of F (x) is √ 2 and the minimum positive period is t = 2 π
X belongs to [0, π] and X + π / 4 belongs to [π / 4,5 π / 4]
It is found that f (x) increases on [π / 4, π / 2] and decreases on [π / 2,5 π / 4]

Let f (x) = 2cosx square + 2 radical sign 3sinxcosx-1 (1) find the period and monotone increasing interval of F (x) (2) It shows how the image of F (x) can be obtained by changing the image of y = SiNx

From the angle doubling formula: F (x) = √ 3sin2x + cos2x
From the auxiliary angle formula: F (x) = 2Sin (2x + π / 6)
1、
Period T = 2 π / 2 = π
Increasing range: - π / 2 + 2K π < 2x + π / 6 < π / 2 + 2K π
-2π/3+2kπ<2x<π/3+2kπ
-Therefore, the decreasing interval is (- π / 3 + K π, π / 6 + K π), K ∈ Z
2、
(1) The abscissa of each point in the image of y = SiNx is changed into 1 / 2 of the original, and the ordinate is unchanged, and the image of y = sin2x is obtained;
(2) The image of y = sin2x is shifted to the left by π / 12 units to obtain the image of y = sin (2x + π / 6);
(3) When y = sin (2x + π / 6), the ordinate of each point in the image is changed to 2 times of the original, and the abscissa remains unchanged, and y = 2Sin (2x + π / 6) is obtained;

The known function f (x) = SiNx + √ 3sinxcosx + 2cosx x x ∈ R How can the image of the minimum positive period and monotone increasing interval function f (x) be obtained from the image of function y = sin2x?

So f (x) = 1 + √ 3sinxcosx + cosx can be changed to 3 / 2 + √ 3 / 2Sin (2x) + 1 / 2cos (2x) according to the angle doubling formula. In this case, it is regarded as sin (π / 6), so the final formula is changed to 3 / 2 + sin (2x + π / 6), and the last minimum positive period is π, which is very clear at a glance!

The known function f (x) = 2 √ 3sinxcosx + 2cosx ^ 2-1

The solution f (x) = 2 √ 3sinxcosx + 2cosx ^ 2-1
=√3*2sinxcosx+（2cosx^2-1）
=√3sin2x+cos2x
=2（√3/2sin2x+1/2cos2x)
=2sin（2x+π/6)
T=2π/2=π

The minimum value of the function y = - root 2cosx + root 2sinx is?

Do you know what SiNx + cosx = is?
So that's the answer
sinx+cosx=√2sin（x+π/4）
Under the Sao year, you have to solve it yourself!

(1) y = 1 / 2cosx + (3 under radical) / 2sinx (2) y = SiNx + cosx

Auxiliary angle formula: asinx + bcosx = (a + b) sin (x + arctanb / a)
(1) Y = 1 / 2cosx + (3 under radical) / 2sinx = sin (x + π / 6)  max = 1, Tmin + = 2 π
(2) Y = SiNx + cosx = (2 under the radical) sin (x + π / 4)) νmax = 2 under the radical, Tmin + = 2 π

It is known that f (x) = sin (2x + π / 3) - radical 3sin ^ 2x + sinxcosx + radical 3 / 2 ① Find the minimum positive period of the function; find the minimum value of the function and the value of X at this time; and find the monotone increasing interval of the function

It can be concluded from the meaning of the title
f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2
=sin(2x+π/3)-√3(1/2-1/2cos2x)+1/2sin2x+√3/2
=2sin(2x+π/3)
(1) Minimum positive period 2 π / 2 = π
(2) The minimum value of the function is - 2, where x = k π - 5 π / 12
(3) The monotone increasing interval of the function is [K π - 5 π / 12, K π + π / 12]

Find the maximum and minimum values of the function f (x) = 2cos 2 (x + π / 6) + √ 3 sin2x, and point out its monotone interval

f(x)=2cos(x+π/6)-1+1+√3sin2x
=cos(2x+π/3)+√3sin2x+1=
=[(1/2)cos2x-(√3/2)sin2x]+)+√3sin2x+1
=[(√3/2)sin2x＋(1/2)cos2x]+1
=sin(2x+π/6)+1
F (max) = 2
F (minimum) = 0
By substituting 2x + π / 6 into the monotone increasing interval of the standard function, the monotone increasing interval of the function is solved as follows:
－π/2+2kπ≤2x+π/6≤π/2+2kπ
－π/3+kπ≤x≤π/6+kπ
The monotone increasing interval of the original function is: [- π / 3 + K π, π / 6 + K π]
Substituting 2x + π / 6 into the monotone decreasing interval of the standard function, the monotone decreasing interval of the function is solved as follows:
π/2+2kπ≤2x+π/6≤3π/2+2kπ
π/6+kπ≤x≤2π/3+kπ
The monotone decreasing interval of the original function is: [π / 6 + K π, 2 π / 3 + K π]

The maximum value of the function y = sin2x-2cos? X?

y=sin2x-2cos²x
=sin2x-(1+cos2x)
=sin2x-cos2x-1
=√2 (√2/2sin2x-√2/2cos2x)-1
=√2sin(2x+π/4)-1
therefore
Maximum = √ 2-1

Find the value range of function y = 2cos ^ 2x + sin2x / 1 + TaNx, thank you very much! To process Oh!

If so, the answer is as follows: 2cos ^ 2x + sin2x = cos2x + sin2x. From the formula cos2x = [1 - (TaNx) ^ 2] / [1 + (TaNx) ^ 2] and sin2x = 2tanx / [1 + (TaNx) ^ 2], y = (2cos ^ 2x + sin2x) /... Can be obtained by the formula cos2x = [1 - (TaNx) ^ 2] / [1 + (TaNx) ^ 2]