Let f (x) = sin (x + 7 π / 4) + cos (x-3 π / 4), X ∈ R. (1) find the minimum positive period sum of F (x) (2) It is known that cos (β - α) = 4 / 5, cos (β + α) = - 4 / 5, 0

Let f (x) = sin (x + 7 π / 4) + cos (x-3 π / 4), X ∈ R. (1) find the minimum positive period sum of F (x) (2) It is known that cos (β - α) = 4 / 5, cos (β + α) = - 4 / 5, 0

(1) Expansion f (x) = SiN x * cos (7 π / 4) + cos x * sin (7 π / 4) + cos x * cos (7 π / 4) - SiN x * sin (7 π / 4)
=√2 *(sin x-cos x)
=√2 *(sin x+sin (x+π/2))
Sum difference product = 2 √ 2 * sin (x + π / 4) * cos (- π / 4)
=2sin(x+π/4)
The minimum positive period is 2 π and the minimum value is - 2
（2）cos(β-α)=cosβcosα+sinβsinα=4/5
cos(β+α)=cosβcosα-sinβsinα=-4/5
Then cos β cos α = 0
sinβsinα=4/5
Another 0

Given the square of function F X sin + 2 SiNx cosx + 3 cosx, find the least positive period and monotone increasing interval of function f (x)

f(x)=sin^22+2sinxcosx+3cos^2x
=1+2sinxcosx+2cos^2
=sin2x+2cos^2-1+1+1
=sin2x+cos2x+2
=√2(sin2x+π/4)+2
Minimum positive period: T = 2 π / 2 = π
Monotonically increasing:
2x+π/4=-π/2+2kπ （k∈Z）
2x=-3π/4+2kπ
x=-3π/8+kπ
2x+π/4=π/2+2kπ （k∈Z）
2x=π/4+2kπ
x=π/8+kπ
Monotone increasing interval: [- 3 π / 8, π / 8] (K ∈ z)

The function y = sin (x - π) 6) The minimum value of cosx______ .

y=sin（x-π
6）cosx=（
Three
2sinx-1
2cosx）cosx=
Three
2sinxcosx-1
2cos2x
=
Three
4sin2x−1
4（cos2x+1）=1
2sin(2x−π
6)-1
Four
∴y=sin（x-π
6) The minimum value of cosx is: − 1
2−1
4＝−3
Four
So the answer is: - 3
4．

It is known that sin α + cos α = (1 + radical 3) / 2, α∈ (0, π / 4) Find the monotone increasing interval of the function f (x) = sin (x - α) + cosx on X ∈ (0, π)

Given that sin α + cos α = (1 + radical 3) / 2, α ∈ (0, π / 4), the function f (x) = sin (x - α) + cosx on X ∈ (0, π) is solved by the simultaneous solution of sin α + cos α = (1 + √ 3) / 2, α ∈ (0, π / 4) and (sin α) ^ 2 + (COS α) ^ 2 = 1

It is known that the maximum value of the function f (x) = sin (x + π / 6) + sin (x - π / 6) + cosx + A is 1 1. Find the value of constant a 2. Find the set of values of X for which F (x) > = holds

(1) F (x) = sin (x + π / 6) + sin (x - π / 6) + sin (x - π / 6) + cosx + a = root number 3 SiNx + cosx + a = 2sinx + cosx + a = 2Sin (x + π / 6) + a because of the function f (x) = sin (x + X + π / 6) + sin (x - π / 6) + cosx + a max value is 1, so, when and only when sin (x + π / 6) = 1, f (x) gets the maximum value, the solution is a = - 1 (2) (x + π / 6) = 1, when, f (x (x)... 6) = 1, f (x)... X (x)... X (x)... X (x)... 6) = 1, f (x)... X (x + it's a good idea

The known function f (x) = sin (x + π / 6) + sin (x - π / 6) + cosx + a (a belongs to R, a is a constant) (1) Finding the minimum positive period of function f (x) (2) If x belongs to [- π / 2, π / 2], the sum of the maximum and minimum values is √ 3, find the value of A

1、
f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a
=2sinxcosπ/6+cosx+a
=√3sinx+cosx+a
=√[(√3)²+1²]sin(x+z)+a
=2sin(x+z)+a
Where Tanz = 1 / √ 3
So t = 2 π / 1 = 2 π
2、
tanz=1/√3
z=π/6
f(x)=2sin(x+π/6)+a
-π/2

The minimum positive period of the function f (x) = sin square (x + π / 4) - Sin square (x - π / 4) is known

Because (x + π / 4) - (x - π / 4) = π / 2, x + π / 4 = π / 2 + (x - π / 4),
So sin (x + π / 4) = sin [π / 2 - (x - π / 4)] = cos (x - π / 4)
So sin square (x + π / 4) - Sin square (x - π / 4) = cos ^ 2 (x - π / 4) - Sin ^ 2 (x - π / 4) = Cos2 [(x - π / 4)] = cos (2x - π / 2) = sin2x
So the minimum positive period 2 π / 2 = π

The value range of the function y = sin2x-2sinx is y ∈______ .

∵ function y = sin2x-2sinx = (sinx-1) 2-1, - 1 ≤ SiNx ≤ 1,
∴0≤（sinx-1）2≤4，∴-1≤（sinx-1）2-1≤3．
The value range of the function y = sin2x-2sinx is y ∈ [- 1,3]
So the answer is [- 1,3]

Let f (x) = cos (x + 2 / 3 π) + 2cos ^ 2 x / 2, X ∈ R Find: 1) the value range of F (x); 2) Note that the opposite side lengths of the inner angles a, B and C of the triangle ABC are a, B, C respectively. If f (b) = 1, B = 1, C = √ 3, find the value of A

(1) (x) = cos (x + 2 π / 3) + 2cos (x / 2) = - (cosx) / 2 - (√ 3sinx) / 2 + 1 + cosx = 1 - [(√ 3sinx) / 2 - (cosx) / 2] = 1 - [sin (x - π / 6)],, 1 / 2 ≤ f (x) ≤ 3 / 2, the domain [0,2] (K ∈ z) (2) a = 1 or 2, f (b) = 1sin ((B - π / 6) = 0b = π / 6 = 6 according to the cosine Theory: B ^ ^ · ·... (b ^... π / 6) = 0b = π / 6 according to the cosine Theory: B ^............ B ^,...,... According to the cosine Theory: B ^,...,... It's a good idea

The maximum value of the function f (x) = cos ^ 2-2cosx is Get it quickly,

f(x)=(cosx-1)²-1
-1