The value range of the function y = radical 3sinx + cosx + 1 is

The value range of the function y = radical 3sinx + cosx + 1 is

y=√3sinx+cosx+1
=2sin(x+30)+1
The range is [- 1,3]

The value range of the function f (x) = - root sign 3sinx + cosx (x belongs to - pi / 2, PI / 2) is Specifically, the value range of the function f (x) = - radical 3 * SiNx + cosx (x belongs to [- π / 2, π / 2]) is How did the step f (x) = - √ 3sinx + cosx = - 2 [sinxcos π / 6-cosxsin π / 6] come into being? -2 [sinxcos π / 6-cosxsin π / 6] is SiNx / 3

f(x)=-√3sinx+cosx=-2[sinxcosπ/6-cosxsinπ/6]=-2sin(x-π/6)
∵-π/2≤x≤π/2
∴-2π/3≤x-π/6≤π/3
∴-1≤sin(x-π/6)≤√3/2
That is - √ 3 ≤ f (x) ≤ 2
The value range is [√] 2

The function f (x) = cosx radical 3sinx, X belongs to the range of [- 3.14/6,3.14/3]______

F (x) = cosx radical 3sinx
=2(cosπ/3cosx-sinπ/3sinx)
=2cos(x+π/3)
∵x∈[-π/6,π/3]
∴x+π/3∈[π/6,2π/3]
The value range of function f (x) is [√ 3, - 1]

What is the value range of the function y = 3sinx + radical 3 * cosx (- π / 2 "X" π / 2)?

[-3,2√3]

Let f (x) = 2 Molecular root 2cos (2x + 4 molecule PI) + SiNx ^ 2, and find the minimum positive period

f(x)=√2/2*(√2/2cos2x-√2/2sin2x)+(1-cos2x)/2=1-1/2*sin(2x) ,
Therefore, the minimum positive period is 2 π / 2 = π

Let z = siny + F (SiNx + siny), where f is a differentiable function. It is proved that (partial Z / partial x) secx + (partial Z / partial y) SecY = 1

Are you sure the title is OK?
It's z = siny + F (SiNx siny)
If so, the proof is as follows
(D is understood as "partial"
dz/dx=dsiny/dx+df(sinx-siny)/dx
=0+df(sinx+siny)/d(sinx-siny)*d(sinx-siny)/dx
=f'(sinx-siny)*cosx
dz/dy=dsiny/dy+df(sinx-siny)/dy
=cosy+df(sinx-siny)/d(sinx-siny)*d(sinx-siny)/dy
=cosy+f'(sinx-siny)*（-cosy）
（dz/dx)secx+（dz/dy)secy= f'(sinx-siny)+1+f'(sinx-siny)*（-1）
=1
If your topic is correct, I can't do it because of my limited ability

Let z = SiNx + F (siny SiNx), where f is a differentiable function, it is proved that: (partial Z / partial x) cosy + (partial Z / partial y) cosx = cosxcosy The answer is second, mainly the process,

dz/dx=cosx+(dF(siny-sinx)/dx)*(-cosx)
dz/dy=(dF(siny-sinx)/dy)*(cosy)
(dz/dx)cosy+(dz/dy)cosx=[cosx+(dF(siny-sinx)/dx)*(-cosx)]cosy+(dF(siny-sinx)/dy)*(cosy)cosx
=cosxcosy-(dF(siny-sinx)/dx)*(cosxcosy)+(dF(siny-sinx)/dx)*(cosxcosy)
=cosxcosy

It is known that the minimum positive period of the function f (x) = sin (PI Wx) coswx + cos ^ 2wx (W > 0) is pi. (1) find the value of W (2) find the points on the image of the function y = f (x) When the abscissa is shortened to 1 / 2 of the original value and the ordinate remains unchanged, the image of the function y = g (x) is obtained, and the minimum value of the function g (x) in the interval [0, PI / 16] is obtained

f(x)=sinwxcoswx+(cos2wx+1)/2
=sin2wx/2+cos2wx/2+1/2
=Root 2 / 2 * sin (2wx + pi / 4) + 1 / 2
pi=2pi/2w
W=1
F (x) = root 2 / 2 * sin (2x + pi / 4)
G (x) = root 2 / 2 * sin (4x + pi / 4)
0<=x<=pi/6,pi/4<=4x+pi/4<=11pi/12
The minimum value is: root 2 / 2 * sin11pi / 12 = (root 3-1) / 4

The maximum value of the function f (x) = sin ^ 2 (x + pi / 12) + cos ^ 2 (x-pi / 12)

f(x)=sin²(x+π/12)+cos²(x-π/12)=1-cos²(x+π/12)+cos²(x-π/12)=1+[cos(x-π/12)+cos(x+π/12)][cos(x-π/12)-cos(x+π/12)]=1+(cosx cosπ/12 + sinx sinπ/12 + cosx cosπ/12 -sinx sinπ...

The value range of the function y = sin (x + π / 3) - SiNx (x ∈ (0, π / 2)) is a.-2,2 b,-1／2,√3／2 c.1／2,1 d.1／2,√3／2

sin(X+π/3)-sinx
＝√3／2cosx-1/2sinx
=cos(x+π/6)
x∈(0,π／2)
Then x + π / 6 ∈ (π / 6,2 π / 3)
Therefore, the value range of the original formula is (- 1 / 2, √ 3 / 2), and B is selected