Given the square of the function f (x) = sinxcosx + cosx, find the value range of F (x)

Given the square of the function f (x) = sinxcosx + cosx, find the value range of F (x)

f(x)=sinxcosx+cos^2 x
=1/2sin2x+1/2cos2x-1/2
=√2/2(sin2x*√2/2+cos2x*√2/2)-1/2
=√2/2sin(2x+π/4)-1/2
Because - 1

The range of the function y = cos squared x-4sin x

Siny-cos = 4x
y=1-sin²x-4sinx
=-sin²x-4sinx+1
=-(sinx+2)²+5
When SiNx = - 1, the function has a maximum value of 4
When SiNx = 1; the function has a minimum value = - 4
So, the range is [- 4,4]

For the function y = sinxcosx + cos ^ 2-2, find the range of the minimum positive period sum function

y=sinxcosx+cos²x-1/2
=(1/2)·2sinxcosx+(1/2)(2cos²x-1)
=(1/2)sin2x+(1/2)cos2x
=(√2/2)[sin2x·cos(π/4)+cos2x·sin(π/4)]
=(√2/2)sin(2x+π/4)
Minimum positive period T = 2 π / 2 = π
Range [- √ 2 / 2, √ 2 / 2]

Cos X-1 / cos x range X-1 / X

The first range is [negative infinity, 0]
The second value is r
1) Let cos? X = t (0 < T ≤ 1)
The original equation is y = T-1 / T
Y '= 1 + 1 / T? Is always greater than 0
So the image of Y increases monotonically on t ∈ (0,1]
When t = 1, y = 0
So, y ≤ 0
2) Let x 2 = t (0 < T)
So f (x) = T-1 / T
So f '(x) = 1 + 1 / T 2 is always greater than 0
So the image of Y increases monotonically on T > 0
So the range is r

F (x) = cos ^ 2x + cosxsinx (0 < x < π / 2) 1. If f (x) = 1, find the value 2 of X and the range of F (x) f(x)=cos^2x+cosxsinx(0＜x＜π/2) 1. If f (x) = 1, find the value of X 2. Find the range of F (x)

F (x) = cos ^ 2x + cosxsinx = (cos2x + 1) / 2 + 1 / 2sin2x = √ 2 / 2Sin (2x + π / 4) + 1 / 2 ① f (x) = 1, 1 = √ 2 / 2Sin (2x + π / 4) + 1 / 2, √ 2 / 2 = sin (2x + π / 4), 2x + π / 4 = 3 π / 4, x = π / 4 ② range [- √ 2 / 2 + 1 / 2, √ 2 / 2 + 1 / 2]

Find the value range of the function y = - cos ^ 2-4sinx + 6

And then we can get the (2) - (x) interval

The value range of the function f (x) = cos ^ 2x -- 4sinx is PS: cos ^ 2x is the square of cosx

f（x）=cos^2x--4sinx
=1-sin^2x-4sinx
Let SiNx = t
Then - 1 < = T < = 1
The original formula = - T ^ 2-4t + 1 = 5 - (T + 2) ^ 2
That is, the range is [- 4,4]

The range of y = 6-4sinx cos ^ 2 x

cosx²=1-sinx²
Suppose k = SiNx
therefore
y=6-4k-1+k²
=k²-4k+5
=（k-2）²+1
Because the range of K is - 1 to 1
So the range of Y is between 10 and 2
So the range is [2,10]

The range of y = cos 2 x-4sinx

y=cos^2x-4sinx
=(1-sin^2x)-4sinx
=1-sin^2x-4sinx
=1-(sinx+2)^2+4
=5-(sinx+2)^2
-1≤sinx≤1
1≤sinx+2≤3
-9≤-(sinx+2)^2≤-1
-4≤5-(sinx+2)^2≤4
Range [- 4,4]

The range of y = - cos ^ 2-4sinx + 5 / 4

If it's not wrong, it's a closed range of minus 11 / 4 to 21 / 4