Given the function y = - sinx-cos2x, then the value range of this function is______ .

Given the function y = - sinx-cos2x, then the value range of this function is______ .

y=-cos2x-sinx=-1+sin2x-sinx=（sinx-1
2）2-5
4，
Because SiNx ∈ [- 1, 1],
So when SiNx = - 1, the maximum value of Y is 1;
When SiNx = 1
At 2, the minimum value of Y is - 5
4，
So the range of the function y is [- 5
4，1]．
So the answer is: [- 5
4，1]．

Find the value range of the function y = 1-sinx + cos ^ 2x O

y=-sin^2x-sinx+2
sinx=[-1,1]
When SiNx = 1, y min = 0
When SiNx = - 1 / 2, ymax = 9 / 4
So the value range
[0,9/4]

The maximum value of the function f (x) = root 3sinx cos ^ 2 is

F (x) = Radix 3sinx-1 + (SiNx) ^ 2, consider SiNx as a whole, let SiNx = t, - 1

The definition domain of the function f (x) = radical 3sinx + cos (x + Θ) is r, and the maximum value is 1. Find the value of Θ The definition domain of the function f (x) = radical 3sinx + cos (x + Θ) is r, and the maximum value is 1. Find the value of Θ

f(x)=√3sinx+cos(x+Θ)
=(√3-sinΘ)sinx+cosΘcosx
Because the maximum value of F (x) is 1, it can be reduced to f (x) = sin (x + α)
Then (√ 3-sin Θ) ^ 2 + (COS Θ) ^ 2 = 1 sin Θ = √ 3 / 2
Θ = π / 3 + 2 π K ∪ 2 π / 3 + 2 π K (k belongs to integer)

F (x) = the maximum value of root 3sinx cos ^ 2x?

F (x) = radical 3sinx cos ^ 2x
=√3sinx-(1-sin^2 x)
=sin^2 x+√3sinx-1
=(sinx+√3/2)^2-7/4
sinx=1
f max=√3

The sum of the maximum and minimum values of the function f (x) = cos ^ 2 + 3sinx / 2 + A is 1 / 8. Find the value of A

Maximum 4 + A
Minimum - 2 + A
a=-15/8

Find the minimum value of the maximum value of the function y = cos ^ x + 3sinx-1 on the closed interval [0,5 π / 6]

Cos ^ x = 1-sin ^ x, y = 1-sin ^ x + 3sinx-1 = - (sin ^ x-3sinx) = - (sinx-1) ^ + SiNx + 1, the maximum value should be (sinx-1) ^ when the minimum is 0, that is, x = π / 2, y = 2,

The minimum positive period of the function y = | cos (2x - (π / 3) |is__ ?

It's directly (2 π / 2) / 2 = π / 2
Because the period of COS (2x - (π / 3) is π, plus the absolute value, that is to flip up the negative half axis of the Y axis, so the period will be halved

Find the minimum positive period of the function y = cos (- 2x + π / 6)

y=cos(-2x+π/6)
=cos(2x-π/6)
T=(2π)/2=π

What is the period of the function y = cos (2x - π / 3)?

The minimum positive period is t = 2 π / 2 = π