Let f (x) = 2Sin ^ 2 (π / 4-x) - radical 3 (sin ^ 2x cos ^ 2x), find the minimum period and monotone decreasing interval of F (x)

Let f (x) = 2Sin ^ 2 (π / 4-x) - radical 3 (sin ^ 2x cos ^ 2x), find the minimum period and monotone decreasing interval of F (x)

f(x)=2sin²（π/4-x）-√3(sin²x-cos²x)
=2[√2／2sinx-√2／2cosx]²-√3cos2x
=(sinx-cosx)²-√3cos2x
=1-sin2x-√3cos2x
=1-2sin(2x+π/3)
So the minimum period T = π, the minimum value is - 1

Find the monotone decreasing interval and symmetry axis of the function FX = sin (2x - π / 4) - 2 radical sign 2Sin? X

fx=sin(2x-π/4)-2√2sin²x=√2/2sin2x-√2/2cos2x-√2(1-cos2x)=√2/2sin2x+√2/2cos2x-√2=sin(2x+π/4)-√2
2kπ+π/2

The function f (x) with respect to X is known to be radical 2Sin (2x + φ) (- π)

According to the image of sine function, the general formula of symmetric axis of sine function is 2x + φ = π / 2 + K π
Take x = π / 8 to obtain φ = π / 4 + K π
By - π

How to deduce y = radical 2 × sin (x + π / 4)

y=sinx+cosx
=√2*(√2/2sinx+√2/2cosx)
=√2*[sinxcos(π/4)+sin(π/4)cosx]
=√2*sin(x+π/4)

Find the value range of function y = radical 3 SiNx / 2-cosx

y=√7/2[sinx*√21/7-cosx*2√7/7)
Let cos θ = √ 21 / 7, sin θ = 2 √ 7 / 7),
y=√7/2sin(x-θ),
Range: y ∈ [- √ 7 / 2, √ 7 / 2]

The known function f (x) = sinx+ cosx，(0≤x≤π 2) Then the value range of F (x) is___ .

From the meaning of the title, f (x)=
sinx+
cosx，(0≤x≤π
2) Two sides squared,
f2(x)=sinx+cosx+2
sinxcosx=
2sin(x+π
4) +
2sin2x，
∵0≤x≤π
2，
ν when x = π
When 4, the function F2 (x) takes the maximum value of 2
2=23
2. When x = 0, the minimum value of F2 (x) is 1,
∵0≤x≤π
2，
∴f(x)=
sinx+
cosx>0，
The value range of F (x) is [1, 23
4]．
So the answer is: [1,23
4]．

F (x) = (1 + SiNx + cosx) [sin (x / 2) - cos (x / 2)] / radical (2 + 2cosx) (1) The simplest form of F (x) (2) Let g (x) = f (x) ^ 2, and make the image of G (x).

(1)-cosx
g(x)=cos^2x=0.5+0.5cos2x

Let f (x) = sin ^ 4-2 radical sign 3sinxcosx cos ^ 4x + 1 When x ∈ [0, π / 3], find the value range of F (x) and the set of X when the maximum value is obtained

F (x) = sin ^ 4x-2 radical 3sinxcosx cos ^ 4x + 1
=(sin ^ 2x cos ^ 2x) (COS ^ 2x + sin ^ 2x) - 2 radical 3sinxcosx + 1
=-Cos2x radical 3sin2x + 1
= -2sin(2x+π/6)+1
When 0

It is known that the function f (x) = 2 radical sign 3sinxcosx cos ^ 2 x + sin ^ 2 x (1) find the minimum positive period of function f (x), (2) when x belongs to [0, π / 2], find the value range of F (x)

If f (x) = 2 radical 3sinxcosx-cos ^ 2 x + sin ^ 2 x = root 3sin2x-cos2 x = 2Sin (2x - π / 6), then the minimum positive period of the function f (x) = t = 2 π / 2 = π 0 ≤ x ≤ π / 2, - 1 / 6 π ≤ 2x - π / 6 ≤ 5 π / 6, then when 2x - π / 6 = - 1 / 6 π, the minimum value of F (x) is - 12x - π / 6 = π / 2

The known function f (x) = 1+ 2sin（2x-π 4）． (1) Find the minimum positive period and maximum value of the function; (2) Find the increasing interval of the function; (3) How can the image of a function be obtained from the image of function y = SiNx?

(1) From the analytic expression of the function f (x), the minimum positive period is 2 π
2 = π with a maximum of 1+
2．
(2) Function f (x) = 1+
2sin（2x-π
4) Monotone interval and function y = sin (2x - π)
4) The monotone interval is the same
Let 2K π - π
2≤2x-π
4≤2kπ+π
2, K ∈ Z, K π - π
8≤x≤kπ+3π
8，
So the increasing interval is [K π - π
8，kπ+3π
8]，k∈z．
(3) The image of y = SiNx is first shifted to the right π
4 unit length, and then shorten the abscissa to the original 1
2 (the ordinate remains unchanged), and then the ordinate is extended to the original one
2 times (the abscissa remains unchanged),
Then, by moving up one unit length, f (x) = 1+
2sin（2x-π
4) The image of