Known function f (x) = sin (π)- δ x)cos δ x+(cos δ x)^2( δ> The minimum positive period of 0) is π Known function f (x) = sin (π)- δ x)cos δ x+(cos δ x)^2( δ> The minimum positive period of 0) is π (1) Beg δ Value of (2) The abscissa of each point on the image of function y = f (x) is shortened to the original 1 / 2, and the ordinate is unchanged. The image of function y = g (x) is obtained, and the minimum value of function g (x) in the interval [0, π / 16] is obtained The answer is (1) 1 （2） 1

Known function f (x) = sin (π)- δ x)cos δ x+(cos δ x)^2( δ> The minimum positive period of 0) is π Known function f (x) = sin (π)- δ x)cos δ x+(cos δ x)^2( δ> The minimum positive period of 0) is π (1) Beg δ Value of (2) The abscissa of each point on the image of function y = f (x) is shortened to the original 1 / 2, and the ordinate is unchanged. The image of function y = g (x) is obtained, and the minimum value of function g (x) in the interval [0, π / 16] is obtained The answer is (1) 1 （2） 1

(1) f(x)=sin(π- δ x)cos δ x+(cos δ x)^2
=sin( δ x)cos δ x+(cos δ x)^2
=(1/2)sin2 δ x+(1+cos2 δ x)/2
=(√2/2)[(√2/2)sin2 δ x+(√2/2)cos2 δ x] +1/2
=(√2/2)sin(2 δ x+π/4)+1/2
So period T = 2 π / (2) δ)= π. Solution δ= one
f(x)=(√2/2)sin(2x+π/4)+1/2
(2) The abscissa of each point on the image of function y = f (x) is shortened to 1 / 2 of the original, and the ordinate is unchanged
g(x)=(√2/2)sin(4x+π/4)+1/2
Let - π / 2 ≤ 4x + π / 4 ≤ π / 2, get - 3 π / 16 ≤ x ≤ π / 16
Therefore, G (x) is an increasing function on [0, π / 16], and the minimum value is g (0) = (√ 2 / 2) sin (π / 4) + 1 / 2 = 1 / 2 + 1 / 2 = 1

Known function f (x) = cos (2pai-x) cos (PAI / 2-x) - Sin ^ 2x Find the minimum positive period and monotone increasing interval of function f (x)

f(x)=cos(2pai-x)cos(pai/2-x)-sin^2x =cosxsinx-sin ² X = (1 / 2) sin2x - (1-cos2x) / 2 = (1 / 2) (sin2x + cos2x) - 1 / 2 = (√ 2 / 2) sin (2x + π / 4) - 1 / 2 the minimum positive period is 2 π / 2 = π when 2x + π / 4 ∈ [2K π - π / 2,2k π + π

[senior one mathematics] known function f (x) = (COS ^ 2) x / 2 - (sin ^ 2) x / 2 + SiNx When x0 ∈ (0, π / 4) and f (x0) = 4 √ 2 / 5, find the value of F (x0 + π / 6)

F (x) = (COS ^ 2) x / 2 - (sin ^ 2) x / 2 + SiNx = (Cos2 * x / 2) + SiNx = SiNx + cosx = √ 2Sin (x + π / 4) f (x0) = √ 2Sin (x0 + π / 4) = 4 √ 2 / 5sin (x + π / 4) = 4 / 5x0 ∈ (0, π / 4) then (x0 + π / 4) ∈ (π / 4, π / 2) so cos (x0 + π / 4) > 0cos (x0 + π / 4) = 3 / 5 get f (x0 + π / 6) =

Known function f (x) = sin ω x( ω> 0). Known function f (x) = sin ω x( ω> 0). 1. When ω= 1, write the analytical function corresponding to the image obtained by translating π / 6 unit length from the image of y = f (x) to the right 2. If y = f (x) image passes through (2 π / 3,0) point and is an increasing function in the interval (0, π / 3), find ω Value of

1.sin(x-π/6)
2. It can be seen that the half period is 2 π / 3, and it is an increasing function in the interval (0, π / 3), so ω＞ 0,2π/ ω＝ 4 π / 3 ω＝ one point five

Given the function f (x) = sin square (x) + (√ 3) sin (x) cos (x) + 2cos square (x), find the minimum positive period and monotonic increasing interval of function f (x). Thank you

f(x)=sin ² (x)+(√3)sin(x)cos(x)+2cos ² (x)
=3/2+√3/2sin2x+1/2cos2x
=3/2+sin(2x+π/6)
The minimum positive period of function f (x) t = 2 π / 2 = π
2kπ-π/2≤ 2x+π/6≤2kπ+π/2,k∈Z
kπ-π/3≤ x≤kπ+π/6,k∈Z
The monotonically increasing interval of F (x) is
[kπ-π/3,kπ+π/6】k∈Z

Given the function f (x) = 2cos (x + π / 3) [sin (x + π / 3) - cos (x + π / 3) with a change number of 3 times], find the value range and minimum positive period of F (x)

f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
=4cos(x+π/3)[1/2 sin(x+π/3)-√3/2 cos(x+π/3)]
=4cos(x+π/3)sin(x+π/3+π/3)
=4cos(x+π/3)sin(x+2π/3)
=4 × (1/2) × [sin(x+2π/3+x+π/3)+sin(x+2π/3-x-π/3)]
=2 × [sin(2x+π)+sin(π/3)]
=2 × [sin(2x+π)+√3/2]
=-2sin(2x)+√3
Value range [√ 3-2, √ 3 + 2]
w=π