Given the function f (x) = 1 / (SiNx + cosx) ^ 2, the function g (x) = 2 + 2sin2x cos ^ 2 (2x), find the definition domain of function f (x) and the value domain of function g (x)

Given the function f (x) = 1 / (SiNx + cosx) ^ 2, the function g (x) = 2 + 2sin2x cos ^ 2 (2x), find the definition domain of function f (x) and the value domain of function g (x)

The definition field of F (x): the denominator is not zero, (SiNx + cosx) ^ 2 = 1 + sin2x ≠ 0  2x ≠ 1.5 π + 2K π, X ≠ 0.75 π + K π
The range of G (x): G (x) = 1 + 2sin2x-sin ^ 2 (2x) [- 2,2]

The function f (x) = a (SiNx + cosx) = B, if a

F (x) = a (SiNx + cosx) = B? = B, right
If + B
We can change a (SiNx + cosx) into asin (x + beat / 4)
A = 1-radical 2,
B = 5-radical 2

To find the maximum and minimum of the function y = SiNx / 2 (SiNx / 2-cosx / 2)

Y = Sin & # 178; (x / 2) - sin (x / 2) cos (x / 2) = (1-cosx) / 2 - (1 / 2) SiNx = 1 - (1 / 2) (SiNx + cosx) = 1 - (√ 2 / 2) sin (x + π / 4) when sin (x + π / 4) = 1, y has the maximum value, y (max) = 1 + √ 2 / 2; when sin (x + π / 4) = 1, y has the minimum value, y (min) = 1 - √ 2 / 2; when sin (x + π / 4) = 1, y has the minimum value, y (min) = 1 - √ 2 / 2;

The minimum value of F (x) = 1 / (SiNx) ^ 2 + 2 / (cosx) ^ 2

f(x)=1*(1/sin²x+2/cos²x)
=(sin²x+cos²x)(1/sin²x+2/cos²x)
=3+2sin²x/cos²x+cos²x/sin²x
≥3+2√2.
If and only if cos & # 178; X = √ 2Sin & # 178; X, the equal sign holds

The minimum value of y = (3 + SiNx) / (1 + cosx)

y+ycosx=3+sinx
sinx-ycosx=y-3
√(1+y^2)sin(x+t)=y-3
|sin(x+t)|=|(y-3)/√(1+y^2)|

The minimum value of y = (SiNx + √ 3) / (cosx + 1) is?

3 / 3, you move the denominator and merge it,

Given the function y = 2cos ^ 2 · x-2sinxcosx, find the period of the function

y=cos2x-sin2x+1,π

Given the function 1 / 2cos ^ 2x + √ 3 / 2sinxcosx + 1, X belongs to R
(1) When the function y takes the maximum, find the set of independent variables X
(2) What kind of translation and expansion transformation can the function image be obtained from the image of y = SiNx（

1/2cos^2x+√3/2sinxcosx+1
=（1/2cos^2x+1）+√3/4sin2x
=1/4cos 2x+√3/4sin2x+5/4
=1/2(sin (2x+π/6))+5/4
1) When the function y takes the maximum value, find the set of independent variables X
That is sin (2x + π / 6) = 1,
2x+π/6=π/2+2kπ(k∈Z)
2x=π/3+2kπ(k∈Z)
X = π / 6 + K π (K ∈ z), where y = 1 / 2 + 5 / 4 = 7 / 4
When the maximum value of function y is 7 / 4, X ∈ {x | π / 6 + K π, K ∈ Z}
(2) What kind of translation and expansion transformation can the function image be obtained from the image of y = SiNx（
Y = sin (x + π / 6) is obtained by moving the y = SiNx image along the X axis to the left by π / 6 units, keeping the ordinate unchanged. The abscissa of y = sin (x + π / 6) image is reduced to 1 / 2 of the original, and the Y = sin (2x + π / 6) image is obtained. Keeping the abscissa unchanged, the total coordinate of y = sin (2x + π / 6) image is reduced to 1 / 2 of the original
1 / 2 (sin (2x + π / 6)) image, the 1 / 2 (sin (2x + π / 6)) image, along the y-axis up translation 5 / 4 units, then get 1 / 2 (sin (2x + π / 6)) + 5 / 4 image, for the Title

Let f (x) = 2sinxcosx. - 2cos ^ 2x (x ∈ R). 1 find the minimum positive period of F (x). 2. When x ∈ [0, Wu / 2], find the value range of the function

F (x) = 2sinxcosx-2cos ^ 2x = sin (2x) - cos (2x) - 1 = √ 2 sin (2x - π / 4) - 1, minimum positive period Tmin = 2 π / 2 = π x ∈ [0, π / 2], then 2x - π / 4 ∈ [- π / 4,3 π / 4] sin (2x - π / 4) ∈ [- √ 2 / 2,1] √ 2 (- √ 2 / 2) - 1 ≤ f (x) ≤ √ 2 - 1-2 ≤ f (x) ≤ √ 2 - 1

The known function f (x) = 2sinxcosx-2cos ^ 2x (x belongs to R)
(1) Finding the minimum positive period of function f (x)
(2) When x belongs to [0,2], find the value range of function f (x)

f(x)=sin2x－cos2x－1
. =√2sin(2x－π/4)－1
The minimum positive period is 2 π / 2 = π
If x ∈ [0, π / 2], then:
2x－π/4∈[－π/4,3π/4]
Then:
sin(2x－π/4)∈[－√2/2,1]
Then: F (x) ∈ [- 2, √ 2-1]