What are the minimum and maximum values of y = cos2x + 2sinx?

What are the minimum and maximum values of y = cos2x + 2sinx?

cos2x=1-2sinx*sinx
Original formula = 1-2sinx * SiNx + 2sinx
=-2(sinx-0.5)^2+1
-1≤sinx≤1
So the maximum is 1
The minimum is - 3.5

The maximum value of the function y = cos2x + 2sinx is______ ．

∵ function y = cos2x + 2sinx = - 2sin2x + 2sinx + 1 = - 2 (SiNx − 12) 2 + 32, ∵ when SiNx = 12, the maximum value of function y is 32, so the answer is: 32

The square function of sincox + y is known
To find the maximum and minimum value of the function, the urgent need for the process, I know it is very troublesome, but I have urgent need, the score is not enough, I hope there are good people can help me! Thank you!

y=sin²x+cos²x+sin2x+2cos²x
=1+sin2x+cos2x+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
-1

f(x)=2sinxcos(3π/2+x)+√3sin(π+x)*cosx+sin(π/2-x)*cosx-1/2
(1) Finding the minimum positive period sum range
(2) Write the set of X when the function f (x) reaches the maximum value
(3) Finding monotone increasing interval of F (x)

(1)
f(x)=2sinxsinx+√3(-sinx)cosx+cosxcosx-1/2
=(1-cos2x)-√3/2sin2x+1/2(1+cos2x)-1/2
=1-1/2cos2x-√3/2sin2x+1/2-1/2
=1-sin(2x+π/6)
T=2π/2=π
y(MAX)=2
y(min)=0
The range is;
[0,2]
(2)
When sin (2x + π / 6) = - 1
When 2x + π / 6 = - π / 2 + 2K π = = > x = - π / 3 + K π, the function takes the maximum,
Set of maximum values:
{x| x= - π/3+kπ,k∈z}
When
When sin (2x + π / 6) = 1
When 2x + π / 6 = π / 2 + 2K π = = > x = π / 6 + K π, the function takes the minimum value,
Set of minimum values:
{x| x=π/6+kπ,k∈z}
(3)
The process of substituting the intermediate variable: 2x + π / 6 into the monotone decreasing interval of the standard function to solve the monotone increasing interval is as follows:
π/2+2kπ≤2x+π/6≤3π/2+2kπ
π/6+kπ≤ x ≤2π/3+kπ
So the interval of the original function is:
[π/6+kπ ,2π/3+kπ]

Simplification (2 (cosx) ^ 4-2 (cosx) ^ 2 + 1 / 2) / (2tan (45 degrees - x) * sin (45 degrees + x) ^ 2))

Molecule = (1 / 2) * [2 (cosx) ^ 2-1] ^ 2 = cos2x / 2
sin(45+x)=cos(45-x)
Denominator = 2Sin (45-x) * cos (45-x) = sin (90-2x) = cos (2x)
Numerator / denominator = 1 / 2

Simplification (2cosa ^ 2-1) / (2tan (π / 4-A) * sin (π / 4 + a) ^ 2)

sin(π/4+a)=cos[π/2-(π/4+a)]=cos(π/4-a) sin(π/4-a)·sin(π/4+a)=(sinπ/4cosa-sinacosπ/4)(sinπ/4cosa+sinacosπ/4)=(cosa·√2/2-sina·√2/2)(cosa·√2/2+sina·√2/2)=cos²a·1/2-sin²a·1/...

Sin & # x-sinxcosx simplify!

sin²x = (1 - cos2x)/2
sinxcosx = 1/2 * sin2x
sin²X-sinXcosX = 1/2 - (cos2x + sin2x)/2 = 1/2 - √2/2 * sin(2x+45°)

The reduction of F (x) = Sin & # 178; ω x + sin ω xcos ω x

From the angle doubling formula of trigonometric function, it can be simplified as:
f(x)=sin²ωx+sinωxcosωx
=1/2(1-cos2ωx)+1/2*2sinωxcosωx
=1/2sin2ωx-1/2cos2ωx+1/2
=√2/2sin(2ωx-π/4)+1/2

Simplification: (1) three SiNx + cosx (2) SiNx cosx (3) 3sinx + 4cosx (4) two sin (π / 4-x)
Simplification: (1) three SiNx + cosx (2) SiNx cosx (3) 3sinx + 4cosx (4) two sin (π / 4-x)

1）2sin（x+π/6）
2) Radical 2Sin (x - π / 4)
3）5sin（x+arccos3/5）
4）1/2（cosx-sinx）