The value range of function f (x) = (SiNx * cosx) / (1 + SiNx + cosx) is

The value range of function f (x) = (SiNx * cosx) / (1 + SiNx + cosx) is

∵sinxcosx
=[（sinx+cosx）^2-1]/2
=(1+sinx+cosx)(sinx+cosx-1)/2
∴y=sinxcosx/(1+sinx+cosx)
=(sinx+cosx-1)/2
And 1 + SiNx + cosx ≠ 0, that is, SiNx + cosx ≠ - 1
And SiNx + cosx = √ 2Sin (x + π / 4) ∈ [- √ 2, √ 2]
The value range of y = sinxcosx / (1 + SiNx + cosx) is
[(-√2-1)/2,-1)∪(-1,(√2-1)/2].
reference resources:
Let SiNx + cosx = t belong to [- root 2, root 2] = > T ^ 2 = 1 + 2sinxcosx = > sinxcosx = (T ^ 2-1) / 2
F (x) = (T ^ 2-1) / 2 (1 + T) = (t-1) / 2 belongs to [- (√ 2 + 1) / 2, (√ 2-1) / 2]
In addition, the denominator is not zero, so 1 + SiNx + cosxb does not = 0, neither t = - 1
To sum up, the value range belongs to [- (√ 2 + 1) / 2, - 1) and up (- 1, (√ 2-1) / 2]

The value range of the function f (x) = SiNx * cosx + SiNx + cosx is

Let SiNx + cosx = t, then the squares of both sides get 1 + 2sinxcosx = T ^ 2,
So f (x) = (T ^ 2-1) / 2 + T = 1 / 2 * (T + 1) ^ 2-1,
From t = SiNx + cosx = √ 2 * sin (x + π) ∈ [- √ 2, √ 2]
When t = - 1, the minimum value of F (x) is - 1,
When t = √ 2, the maximum value of F (x) is √ 2 + 1 / 2,
Therefore, the value range of the function is [- 1, √ 2 + 1 / 2]

The range of the function f (x) = SiNx / (2-cosx) is

Let SiNx / (2-cosx) = t
sinx=2t-tcosx
sinx+tcox=2t
Root sign (T ^ 2 + 1) sin (x)+ ξ)= 2t （ ξ (auxiliary angle)
sin(x+ ξ)= 2T / root sign (T ^ 2 + 1) - 1

The value range of function y = (1 + SiNx) / (2 + cosx) is

2y+ycosx=1+sinxsinx-ycosx=2y-1√(1+y ²) sin(x+ θ)= 2y-1 then: sin (x)+ θ)= (2y-1)/√(1+y ²) Obviously: - 1 ≤ (2y-1) / √ (1 + y) ²) ≤ 1, i.e. (2y-1) ²/ (1+y ²) ≦1 (2y-1) ² ≦y ²+ 1 3y
...

Given the function y = sin2x + SiNx + cosx + 2, find the value range

Let t = SiNx + cosx = √ sin (2x + π / 4), so t belongs to [- √ 2, √ 2] sin2x = 2sinxcosx = (SiNx + cosx) ^ 2-1, so y = t + (t * t-1) / 2 + 2 = 0.5T * t + T + 1.5, t belongs to [- √ 2, √ 2] when x = √ 2, y has a maximum value of 2.5 + √ 2x = - 1, and a minimum value of 1, so the y value field is [1,2.5 + √ 2]

Function y = 1 + SiNx / 2 + cosx, find the value range,

If the title means: y = 1 + sin (x / 2) + cosx
So there
y＝1+sin（x/2）+cosx=1+sin（x/2）+1 − 2sin^2 （x/2)
Let t = sin (x / 2) where t belongs to [- 1,1]
Then y = - 2T ^ 2 + T + 2 T belongs to [- 1,1]
Then it is solved according to the quadratic function