Given the function f (x) = sinxcosx - √ (3) sin? X, find the maximum and minimum value of the minimum positive period of (1) FX and (2) FX on [0, π / 2]

Given the function f (x) = sinxcosx - √ (3) sin? X, find the maximum and minimum value of the minimum positive period of (1) FX and (2) FX on [0, π / 2]

F (x) = - 3 / 2 * (cos2x-1) + 1 / 2 sin2x (cos2x-1) + 1 / 2sin22x = √ 3 / 2cos2x + 1 / 2sin22x - √ 3 / 2 = cos (2x - π / 6) - √ 3 / 2, the minimum positive period T = 2 π / 2 = π, 2, X ∈ [0, π / 2], then 2x - π / 6 ∈ [- π / 6,5 π / 6] when 2x - π / 6 = 0, f (x (x x x x x x x x x = 0, X ∈ [0, π / 2], then 2x - π / 6 ∈ [- π / 6 ∈ [- π / 6,5 π / 6] when 2x take the maximum value of 1 - √ 3 / 2

Given the function y = sin ^ 2x-1 / 2sinx + 1, if y takes the maximum value, x = α, and Y takes the minimum value, x = β, and α, β belong to If - π / 2, π / 2], then sin (α - β) =?

Let SiNx be t, where t is [- 1,1], then the original formula is y = (t-1 / 4) ^ 2 + 15 / 16, if y is the largest, then t = - 1, that is, Sina = - 1, cosa = 0, if y is the smallest, then t = 1 / 4, that is, SINB = 1 / 4, CoSb = √ 15 / 16, so sin (a-b) = sinacosb cosasinb = - 1 ×√ 15 / 16 - 0 × 1 / 4 = - √ 15 /

The minimum and maximum of the function f (x) = cos? X-sin? X + 2sinx, X ∈ R

f(x)=cos²x-sin²x+2sinx
f(x)=1-sin²x-sin²x+2sinx
=1-2sin²x+2sinx
Let t = SiNx, then the range of T is [- 1,1]
g(t)=1-2t²+2t
The axis of symmetry is t = 1 / 2
So the maximum value is t = 1 / 2 g (1 / 2) = 1-2 * (1 / 2) 2 + 2 / 2 = 3 / 2
So the minimum value is t = - 1 g (1 / 2) = 1-2 * (- 1) 2 + 2 (- 1) = - 3

The maximum and minimum of the function y = sin 2 X-5 / 2sinx + 5 / 2 Seeking process

formula
y=(sinx-5/4)²+15/16
-1<=sinx<=1
On the left side of the axis of symmetry SiNx = 5 / 4
Decline
therefore
SiNx = - 1, the maximum value is 4
SiNx = 1, the minimum is 1

The maximum value of the function f (x) = cosx sin? X-cos2x + 7 / 4 is——

f(x)=cosx-sin²x-cos(2x) +7/4
=cosx-(1-cos²x)-(2cos²x-1)+7/4
=-cos²x+cosx+7/4
=-(cosx -1/2)² +2
-When cosx = 1 / 2, f (x) has a maximum value f (x) max = 2

F (x) = sin (2x + π / 6) Let G (x) = f (x) - cos2x, find the maximum and minimum value of function g (x) on the interval x ∈ [0, π / 2]

 

Given the function y = cos2x + sin ^ 2-cosx. Find the maximum and minimum values

cos2x=cos^2-sin^2
cos2x+sin^2-cosx
=cos^2-sin^2+sin^2-cosx
=cos^2-cosx
=cos^2-cosx+1/4-1/4
=(cosx-1/2)^2-1/4
-1

(1) What is the minimum value of the function y = cosx sin ^ 2x-cos2x + 17 / 4 (2) What is the even function with the smallest positive period π? A.y=sin2x B.y=cosx/2 C.y=sin2x+cos2x D.y=1-tan^2/1+tan^2 (3) Let θ be the second quadrant angle, then there must be () A.tanθ/2>cotθ/2 B.tanθ/2cosθ/2 D.sinθ/2

Y = cosx + sin ^ 2x-cos2x + 17,4 = cosx + 1-cos ^ 2x (2cos ^ 2x-1) + 17,4 = - 3cos ^ 2x + cosx + 25,4 = - 3 (cosx-1,6) ^ 2 + 19,3 when cosx = - 1, ymein = 9,42.da.y = sin2x t = 2 π (2 = 2 = π), but the odd function, so a is not correct. B.y = cosx-2 t = 2 π (1,2) = 4 π (1,2) = 4 π ≠ 4 π ≠ 4 π ≠ 4 π ≠ 4 π, b.y = cosx = cosx-2 t = 2 T = 2 π (1,2) = 4 π (1,2 π so B

(1) The function y = 2sinx-3cosx, (2) y = cos square x-cos4 power X, with respect to their maximum and minimum periods

In the form of rounds, Hao Junmin has played all the Bundesliga matches in the Schalke 04 football team this year

The least positive of the function f (x) = sin quartic x + cos quartic x + sin quadratic xcos quadratic x ratio of 2 - sin2x is obtained Find the minimum positive period, maximum and minimum value of the function f (x) = sin quartic x + cos quartic x + sin quadratic x cos quadratic x ratio,

The minimum positive period: T = Pi, max = 13 / 8, min = - 3 / 8