The value range of the function y = SiNx + cos2x is () A. [-1，5 4] B. [-1，1] C. [1，4 5] D. （-∞，4 5]

The value range of the function y = SiNx + cos2x is () A. [-1，5 4] B. [-1，1] C. [1，4 5] D. （-∞，4 5]

y=sinx+cos2x=sinx+1-sin2x=-（sinx-1
2）2+5
4，
∵sinx∈[-1，1]，
∴sinx=1
2, ymax = 5
When SiNx = - 1, Ymin = - 1,
The value range of the function is [- 1,5
4]．
So choose a

The value range of the function y = cos (x + 5 °) + 3 √ 2cos (x + 50 °),

y=cos（x+5°）+3√2cos（x+5°+45°）
=cos（x+5°）+3√2·[cos（x+5°）cos45°+sin（x+5°）sin45°]
=4cos（x+5°）+3sin（x+5°）
=5 · sin (x + 5 ° + a) (where Tana = 4 / 3)
Therefore, the value range is [- 5, 5]

It is known that the function f (x) = cos (ω x + ψ (ω > 0,0 < ψ < π) is an odd function defined on R, and in the interval [0, π)/

1. From the odd function, f (0) = 0, thus cos φ = 0, therefore, 0 ≤ φ ≤ π, so φ = π / 2
If f (x) is a monotone function in the interval [0, π / 2], w > 0, we can get that w * π / 2 + π / 2 is less than or equal to π, so w is less than or equal to 1
When x = 3 π / 4, f (x) reaches the maximum value, w * 3 π / 4 + π / 2 = k π can be obtained, w = 2 / 3
2. Transform from Triangle: F (x) = - sin (2 / 3x)
(odd to even constant, symbol to see quadrant)
-The maximum value of SiNx can be obtained easily when x = 2K π - π / 2, and can be obtained at 2 / 3 * a = 2008, thus a = 3012

It is known that the function f (x) defined on R satisfies at the same time that ① f (0) = f (45 °) = 1; ② f (M + n) + F (m-n) = 2F (m) cos (2n) + 8sin? X What is the maximum value of function f (x)

The last 8sin, the letter is wrong

Let f (x) = cos (2x + Pai / 3) + (sin ^ 2) x 1: find the range and minimum positive period of F (x). 2: let a, B, C Let me ask you a math problem: Let f (x) = cos (2x + Pai / 3) + (sin ^ 2) X 1: Find the range of F (x) and the minimum positive period 2: Let a, B and C be the three inner angles of △ ABC, and their opposite side lengths are ABC respectively. If cos C = (2 √ 2) / 3, a is an acute angle and f (A / 2) = (- 1 / 4) a + C = 2 + 3 √ 3, calculate the area of ABC

F (x) = cos (2x + Pai / 3) + (SiNx) ^ 2
=(cos2x)/2-√3(sin2x)/2+(1-cos2x)/2
=1/2－√3(sin2x)/2
From - 1 ＜ = sin2x ＜ = 1
The value range of F (x) is [(1 - √ 3) / 2, (1 + √ 3) / 2]
Period T = 2 π / 2 = π
There is no place to write!

Find the value range of the function y = cos ^ 2x-sin ^ x-4sinx + 1, X ∈ R

y=1-2sin²x-4sinx+1
=2-2sin²x-4sinx
Let SiNx = t, then y = - 2T? - 4T + 2, t ∈ [- 1,1]
When the opening is downward, the symmetry axis is t = - 1 and decreases monotonically
When t = 1, Ymin = - 4
When t = - 1, ymax = 4
The range is [- 4,4]

The known function f (x) = 2 (COS ^ 2 × x + √ 3sinxcosx) + 1 1) Find the minimum positive period of F (x) and find its monotone decreasing interval 2) When x ∈ [0, п / 2], find the range of F (x) The Pai of 3.14

① (x) = 2cos * x + 2 √ 3sinxcos x + 1 = 1 + cos2x + √ 3sin2x + 1 = 2Sin (2x + π / 6) + 2T = π single reduction interval: 2K π + π / 2 ≤ 2x + 2x + π / 6 ≤ 2K π + 3 π / 2 (k ∈ z) that is k π + π / 6 ≤ x ≤ K π π + 2 π / 3 (K ∈ z), namely K π + π / 6 ≤ x ≤ K π π + 2 π / 3; ② when x ∈ [0, π / 2], 2x + π / 6 ∈ [π / 6,7 π / 6,7 π / 7 π / 6,7 π / 6,7 πtherefore, when 2x +

Let f (x) = 4cos (ω X - π / 6) sin ω x-cos (2 ω x + π), where ω > 0, find the value range of function y = f (x), please see the supplement of the problem f（x）=4cos（ωx-π/6）sinωx-cos（2ωx+π） =4(coswxcosπ/6+sinwxsinπ6)sinwx+cos2wx =2√3sinwxcoswx+2sin²wx+cos2wx =√3sin2wx+1-cos2wx+cos2wx =√3sin2wx+1 The maximum value is 1 + √ 3, and the minimum value is 1 - √ 3 I don't know how to get the maximum and minimum

U = maximum value of sinwx = 1, minimum value = - 1,
The maximum value of F (x) = √ 3U + 1 = 1 + √ 3, the minimum value = 1 - √ 3
Is that OK?

Let f (x) = 4cos (Wx - π / 6) sinwx cos (2wx + π), where w > 0. Find the value range of the function y = f (x), if y = f (x) is an increasing function on the interval [- 3x / 2, π / 2], the maximum value of W is obtained“

It is expanded as follows: 4 [√ 3 / 2coswx + 1 / 2sinwx] sinwx + (coswx) ^ 2 - (sinwx) ^ 2 = 2 √ 3coswxsinwx + 1 = √ 3sin2wx + 1 [if the definition field of this problem is all real numbers, then the value range is (- √ 3 + 1, √ 3 + 1)] the original title should be in the interval [- 3 π / 2, π / 2], if so, [- 3 π W, π w] ∈ [2K π - π / 2,2

Let f (x) = 4cos (ω X - π / 6) sin ω x-cos (2 ω x + π), where ω > 0, (1) find the function y = F Let f (x) = 4cos (ω X - π / 6) sin ω x-cos (2 ω x + π), where ω > 0 (II) if f (x) is an increasing function in the interval [- 3 π / 2, π / 2], find the maximum value of ω Why is k equal to 0?

This can be done with cos (α - β). You'd better ask the teacher