A monotone increasing interval of the function f {a} = cos square a-2cos square A / 2 Please write down the steps and results,

A monotone increasing interval of the function f {a} = cos square a-2cos square A / 2 Please write down the steps and results,

f{a}=cos?a-cos?a/2
=cos?a-cosa-1
=(cosa-1/2)?-3/2
∵ the axis of symmetry is 1 / 2
When cosa > 1 / 2, it is increased
That is {x | - π / 3 + 2K π

Is the monotone increasing interval of the function f (x) = cos? X-2cos? X / 2?

f(x)=cos²x-2cos²（x/2)
f'(x)=-2sinxcosx+sinx=sinx(1-2cosx)
Monotonic increase
sinx>0,cosx

Let f (x) = cos (x + 2) 3π）+2cos2x 2，x∈R． (1) Find the range of F (x); (2) Note that the opposite side lengths of △ ABC inner angles a, B and C are a, B, C respectively, if f (b) = 1, B = 1, C= 3. Find the value of A

(1) F (x) = cos (x + 23 π) + 2cos2x2 = cosxcos23 π - sinxsin23 π + cosx + 1 = - 12cosx-32sinx + cosx + 1 = sin (x + 5 π 6) + 1, so the value range of F (x) is [0, 2] (II) from F (B) = 1, sin (B + 5 π 6) + 1 = 1, that is sin (B + 5 π 6) + 1 = 1, that is sin (B + 5 π 6) =... "

The known function f (x) = 2cos ^ 2 (ω X / 2) + cos (ω x + π / 3) (ω > 0) this is it

f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx)cosπ/3-sin(ωx)sinπ/3
=1+3cos(ωx)/2-√3sin(ωx)/2
=1+√3*[√3/2cos(ωx)-1/2sin(ωx)]
=1+√3*[sinπ/3cos(ωx)-cosπ/3sin(ωx)]
=1+√3*sin(π/3-ωx)
Or we can use sum difference product directly
f(x)＝2cos^2(ωx/2)+cos(ωx+π/3)
=1+cos(ωx)+cos(ωx+π/3)
=1+2cos[(ωx+ωx+π/3)/2]cos(ωx-ωx-π/3)/2]
=1+2cos(ωx+π/6)cos(π/6)
=1+√3cos(ωx+π/6)

What is the maximum value of the function y = 2sinx squared + 2cos X - 3?

y=2sinx^2+2cosx-3
=2(1-cosx^2)+2cosx-3
=2-2cosx^2+2cosx-3
=-2cosx^2+2cosx-1
=-2(cosx^2-cosx+1/4)-1/2
=-2(cosx-1/2)^2-1/2
So when cosx = 1 / 2, the maximum value is - 1 / 2

Find the function f (x) = 2cos square x + 2sinx_ The maximum value and minimum value of 3, and find the x value of printing

f(x)=2(1-sin²x)+2sinx-3
=-2sin²x+2sinx-1
=-2(sinx-1/2)²-1/2
-1

The minimum positive period and parity of the function y = 2cos 2 (x - π / 4) - 1 And the minimum positive period of F (x) = 4sin 2 (π / 6 + x) I don't know how to square it,

y=2cos²(x-π/4)-1=cos(2x-π/2)=cos(π/2-2x)=sin2x
The minimum positive period π of the function y = 2cos 2 (x - π / 4) - 1 is an odd function
f(x)=4sin²(π/6+x)=1-2x[1-cos(π/3+2x)]=1-2+2cos(π/3+2x)
=2cos(π/3+2x)-1
The minimum positive period π of F (x) = 4sin 2 (π / 6 + x)
This kind of questions use cos2x = cos? X-sin? X = 2cos? X-1 = 1-2sin? X

The minimum positive period of the function y = cos (π x) / 2cos (π / 2 (x-1)) is

y=cos[(πx)/2]cos[π(x-1)/2]=1/2*{cos[(πx)/2+π(x-1)/2]+cos[(πx)/2-π(x-1)/2]
=1/2*[cos(πx-π/2]+cos(π/2)]
=1/2*sin(πx)
Therefore, the minimum positive period is t = 2 π / π = 2

What is the minimum positive period of the function y = 2cos square X-1

The minimum positive period is t = 2 π / 2 = π

The minimum positive period of the function y = 2cos 2 x 1

y=2cos²x
y=1+cos2x
Therefore, it can be found that
The minimum positive period T = 2 π / w = 2 π / 2 = π
(the minimum positive period of cos2x is π, 1 + cos2x only shifts up one unit based on the original image, and does not change the periodicity of the function)