Given the square (x / 2) + sinx-1 of function FX = 2cos, find the minimum positive period sum range of function FX The square (x / 2) + sinx-1 of the known function FX = 2cos Finding the minimum positive period sum range of function FX If x belongs to [π / 2,3 π / 4], and FX = 1 / 5, find the value of SiNx

f(x)=[2cos^2(x/2)-1]+sinx
=cosx+sinx
=√2sin(x+π/4)
∵x∈R ∴x+π/4∈R
∵f(x)=sinx∈(-1,1)
∴f(x)=√2sin(x+π/4)∈(-√2,√2)
T=2π/1=2π

When x ∈ [π / 6,7 π / 6], find the minimum and maximum of the function y = 3-sinx-2cos ^ 2x

sin^2x+cos^2x=1
So f (x) = 3-sinx-2 (1-sin ^ 2x)
=2sin^2x-sinx+1
=2(sinx-1/4)^2+7/8
π/6

When x belongs to the [sixth school, seven sixth school], the minimum value? Maximum value? Of the function y = 3-sinx-2cos square x?

y=3-sin x-2cos^2 x
=3-sin x-2+2sin^2 x
=2sin^2 x-sin x+1
=2(sin x-1/2)^2+1/2
X=[π/6,7π/6]
sinx=[-1/2,1]
therefore
ymin=1/2
ymax=5/2

We know that the function y = (SiNx + cosx) squared + 2cosx squared. (1) find its decreasing interval (2) find its maximum and minimum The detailed process is mainly how to find the definition domain of his decreasing interval

y=sinx^2+cosx^2+2sinxcosx+2cosx^2
= 1+sin2x+1+cos2X
=2 + radical 2 * sin (2x + π / 4)
If it is monotonically decreasing, then 2x + π / 4 ∈ (2k π + π / 2,2k π + 3 * π / 2) is required
The interval comes out
(2) When sin (2x + π / 4) reaches 1, it is 2 + radical 2
When sin (2x + π / 4) reaches - 1, it is 2-radical 2

Let f (x) = the square of cosx-2sinxcosx SiNx Find the minimum positive period! Maximum minimum value! Monotone interval! "= √ 2cos (2x + π / 4)" how did it come from?

Simplification with the formula of double angle
f(x)=cosx^2-2sinxcosx-sinx^2=cos2x-sin2x
= √2cos(2x+π/4)
So the minimum positive period T = 2 π / 2 = π, the maximum value is √ 2, and the minimum value is - √ 2
As for the solution of monotone interval, 2x + π / 4 is regarded as a whole
Let 2K π < 2x + π / 4 < 2K π + π
So K π - π / 8, so the monotone decreasing interval is (K π - π / 8, K π + 3 π / 8), K ∈ Z
The monotone increasing interval is (K π + 3 π / 8, K π + 7 π / 8), K ∈ Z

Let f (x) = the square of cosx-2sinxcosx SiNx, X ∈ [0,2 / π], find the maximum value of F (x)

Simplification with the formula of double angle
f(x)=cosx^2-2sinxcosx-sinx^2=cos2x-sin2x
= √2cos(2x+π/4)
Let 2K π < 2x + π / 4 < 2K π + π
So K π - π / 8, so the monotone decreasing interval is (K π - π / 8, K π + 3 π / 8), K ∈ Z
The monotone increasing interval is (K π + 3 π / 8, K π + 7 π / 8), K ∈ Z
Because x ∈ [0,2 \ \ π]
So f (x) max = f (0) = 1
F (x) min = f (8-3 π) = - √ 2

The known function f (x) = cosx ^ 4-2sinxcosx-sinx ^ 4 (1) Find the minimum positive period of F (x); (2) If x belongs to [0, Pai / 2], find the maximum and minimum of F (x)

Because (sinx-cos4) = SiNx
=(cosx^2+sinx^2)(cosx^2-sinx^2)-sin2x
So the minimum positive period of F (x) t = 2pai / 2 = Pai
Because 0

The known function f (x) = (cosx) ^ 4-2sinxcosx - (SiNx) ^ 4 ① Find the minimum value of F (x) ② If x ∈ [0, π / 2], find the maximum and minimum of F (x)

1、
f(x)=(cos²x+sin²x)(cos²x-sin²x)-2sinxcosx
=1*cos2x-sin2x
=-(sin2x-cos2x
=-√2sin(2x-π/4)
So t = 2 π / 2 = π
2、
Zero

Find the maximum value of the square of the function f (x) = (SiNx) + 2sinxcosx + 3 (cosx), and find the value of X at this time

f(x)=(1-cos2x)/2+sin2x+3(1+cos2x)/2
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
So 2x + π / 4 = 2K π + π / 2
When x = k π + π / 8, the maximum value = √ 2 + 2

Given the vector m (SiNx, cos θ), n (cosx, sin θ), m * n = √ 10 / 10. If θ = π / 8, find sin2x

Because m (SiNx, cos, θ), n (cosx, sin, θ), θ = π / 8
So m * n = sinxcosx + cos θ sin θ
=(1/2)*[sin(2x)+sin(2θ)]
=(1/2)*[sin(2x)+sin(π/4)]
=(1/2)*[sin(2x)+√2/2]
=√10/10
So sin (2x) = 2 * √ 10 / 10 - √ 2 / 2 = (2 √ 10-5 √ 2) / 10