The known function f (x) = root sign 3sinx cosx, X belongs to the range of R to find f (x) SiNx and cosx are not included in root 3,

The known function f (x) = root sign 3sinx cosx, X belongs to the range of R to find f (x) SiNx and cosx are not included in root 3,

F (x) = Radix 3sinx cosx = 2 (√ 3 / 2sinx-1 / 2cosx)
=2sin(x-π/6)
Therefore, the value range [- 2,2]

The maximum value of the function y = 2sinx cosx is______ .

y=2sinx-cosx=
5sin（x+φ）≤
Five
So the answer is:
Five

Find the maximum value of the function y = 2 + cosx The value is 3. Please write the process

y=(2-cosx)/(2+cosx)=4/(2+cosx)-1
-1

The function y = cosx + cos (x + π) 3) The maximum value of is___ .

∵y=cosx+cos（x+π
3）=cosx+1
2cosx-
Three
2sinx
=3
2cosx-
Three
2sinx=
3cos（x+π
6）
So y = cosx + cos (x + π)
3) The maximum value of is
Three
So the answer is:
Three

The maximum value of the function y = cosx = cos (x + Π / 3) is? The process should be detailed After cosx+ Drivers of ar your solution is a little troublesome In addition, I know to use C (α ± β) to solve the problem, but I can't deduce it

y=cosx+cos(x+∏/3)
=Cosx + cosx / SiNx 3 times the root of 2-2
=-(1.5cosx-sinx of 3 times the root of 2 / 2)
=-Radical 3sin (x -?)
The minimum value of sin (x -?) is - 1
-The maximum value of root 3 sin (x -?) = y is root 3
It doesn't matter how much

Find the function y = cosx + cos (x - π 3) The maximum and minimum of (x ∈ R)

∵y=cosx+cos（x-π
3）
=cosx+cosxcosπ
3+sinxsinπ
Three
=3
2cosx+
Three
2sinx
=
3（cosπ
6cosx+sinπ
6sinx）
=
3cos（x-π
6），
∵-1≤cos（x-π
6）≤1，
∴ymax=
3，ymin=-
3．

Known function y = SiNx 2+ 3cosx 2，x∈R． (1) Find the set of X corresponding to the maximum value of Y; (2) The image of y = SiNx (x ∈ R) can be obtained by the translation and extension transformation of the image of this function

y＝sinx
2+
3cosx
2＝2sin(x
2+π
3)
(1) When x
2+π
3＝2kπ+π
2, that is, x = 4K π + π
When k ∈ Z, y gets the maximum value {x | x = 4K π + π
3, K ∈ Z}
（2）y=2sin（x
2+π
3) Shift right 2 π
3 units
y=2sinx
The abscissa is reduced to 1
2 times
The y = 2sinx ordinate is reduced to the original 1
2 times
y=sinx

Given that the function y = radical three SiNx + cosx, X belongs to R, when the function y gets the maximum value, find the set of independent variables X The image of this function can be obtained from the image of y = SiNx (x belongs to R)

Y = radical 3 / 4 * sin2x + 1 / 4 * cos2x + 5 / 4
=1/2*sin(2x+π/6)+5/4
When 2x + π / 6 = 2 / π + 2K π, the maximum value of Y is 7 / 4
In this case, x = π / 6 + K π

The maximum value of the function y = 2 + SiNx cosx

y=2+sinx-cosx
=2+√2(√2/2sinx-√2/2cosx)
=2+√2(sinxcosπ/4-cosxsinπ/4)
=2+√2sin(x-π/4)
So the maximum value is 2 + √ 2

Find the maximum value of the function g (x) = sin ^ 2x + SiNx * cosx

g(x）=sin^2x+sinx*cosx=1/2+1/2sin2x-1/2cos2x=1/2+√2/2sin(2x-π/4)
The maximum value is (1 + √ 2) / 2
The minimum value is (1 - √ 2) / 2