Given the function f (x) = SiNx (1 + SiNx) + cos squared * X. (1) find the maximum and minimum of F (x) on the [negative 6 th, 2 / 3 th] school Given the function f (x) = SiNx (1 + SiNx) + cos squared * X. (1) find the maximum and minimum value of F (x) on the [negative 6 th, 2 / 3 th] school. (2) in the triangle ABC, given cosa = 7 / 25 and CoSb = 3 / 5, find f (c)

f（x）
=SiNx (1 + SiNx) + cos square * x
=（sinx）^2+sinx+(cosx)^2
=1+sinx
(1) F (x) is an increasing function in [minus 6, 2] and decreasing in (2, 2, 3)
F (x) max = f (2 / 2) = 1 + 1 = 2
F (negative 6-th school) = 1 / 2 < 1 + radical 3 / 2 = f (2 / 3 school)
So the minimum value is 1 / 2
(2) In a triangle, each angle is less than 180 degrees
Cosa = 7 / 25, CoSb = 3 / 5
So Sina = 24 / 25, SINB = 4 / 5
SINC=sin(A+B)=SINACOSB+COSASINB=24/25*3/5+7/25*4/5=100/125=4/5
So f (c) = 1 + sinc = 1 + 4 / 5 = 9 / 5

If | x | ≤ π / 4, then the minimum value of the function y = cos ^ 2 x + SiNx is

y = 1 - (sinx)^2 + sinx
= -(sinx-1/2)^2 + 5/4
|X | ≤ π / 4, so - √ 2 / 2

Let 0 < a │≤ 2, the maximum value of the square X of the function f (x) = COS is 0, the minimum value is - 4, and the angle between a and B is 45 °, a. B is a vector

Because (SiNx) ^ 2 + (cosx) ^ 2 = 1
So f (x) = 1 - (SiNx) ^ 2 - | a | SiNx - | B|
Let SiNx = M
Primitive functionalization - m ^ 2 - | a | m + 1 - | B | and - 1

The function f (x) = cos2x + SiNx in the interval [- π 4，π 4] The minimum value on is () A. 2−1 Two B. -1+ Two Two C. -1 D. 1− Two Two

f（x）=cos2x+sinx=1-sin2x+sinx=-（sinx-1
2）2+5
4．
∵x∈[-π
4，π
4] So SiNx ∈ [−
Two
2，
Two
2]
Therefore, when SiNx = -
Two
When 2, the minimum value of the function is Ymin = 1 −
Two
2．
That is, if x = - π
At 4, Ymin = 1 −
Two
2．
Therefore, D

Let f (x) = cos square x + (A-1) SiNx + A, a belongs to R. when a = 2, find the maximum value of function f (x)

When a = 2
F(x)=cos^2x+(a-1)sinx+a
= cos^2x+sinx+2
= 1-sin^2x+sinx+2
=-(sinx-1/2)^2+13/4
-(sinx-1/2)^2 ≤ 0
∴-(sinx-1/2)^2+13/4 ≤ 13/4
13 / 4 Max

Find the maximum and minimum of the square of the function y = cos (9 / 2 π + x) + SiNx

y=cos(9/2π+x)+sinx=cos(π/2+x)+sinx=-sinx+sinx=0=max=min
y=cos(9/(2π)+x)+sinx=cos(9/(2π))cosx-(sin(9/(2π))-1)sinx
=(-√(cos(9/(2π)))^2+(sin(9/(2π))-1)^2)cos(x-c)
tanc=(sin(9/(2π))-1)/cos(9/(2π))
ymin=-√(cos(9/(2π)))^2+(sin(9/(2π))-1)^2
ymax=√(cos(9/(2π)))^2+(sin(9/(2π))-1)^2

Given the function y = (SiNx + COS) squared + 2cos squared x (1) find its decreasing interval (2) find its maximum and minimum values

1 + 1 + sin2x + 2cos ^ 2x-1 + 1 = sin2x + cos2x + 2 = √ 2Sin (2x + π / 4) + 22x + π / 4 = π / 2 + 2K π (K ∈ z) 2x = π / 4 + 2K π (K ∈ z) 2x = π / 4 + 2K π x = π / 8 + 2K π x = π / 8 + 2K π x = π / 8 + 2K π (K ∈ z) 2x = 5 π / 4 + 2K π (K ∈ z) 2x = 5 π / 4 + 2K π x = 5 π / 8 + 8 + k + K π (π) decline interval [π / 8 + 8 + K π5 π / 8 + K π] (K ∈ z) y (max) = √ 2 + 2

Find the maximum value of the function y = (2sinx * cos ^ 2x) / (1 + SiNx), X ∈ [- π / 4, π / 4]

y=(2sinx*cos^2x)/(1+sinx)
=(2sinx(1- sin²x))/(1+sinx)
=2sinx(1-sinx)
=2sinx-2sin²x
Let SiNx = t ∈ [- √ 2 / 2, √ 2 / 2],
y=2t-2t²=-2(t-1/2) ²+1/2.
When t = 1 / 2, the maximum value of the function is 1 / 2
When t = - √ 2 / 2, the minimum value of the function is - √ 2-1

Find the maximum and minimum values of the function y = 7-2sinx cos ^ 2x

y=7-2sinx-cos^2x=7-2sinx-1+sin^2x
Let SiNx = t
The original formula = 7-2t-1 + T? = 6-2t + T? = (t-1) 2 + 5, because: | t | ≤ 1, - 1 ≤ t ≤ 1
So T-1 ∈ [- 2,0]
So (t-1) 2 ∈ [0,4]
So (t-1) 2 + 5 ∈ [5,9]
So the maximum value of this function is 9 and the minimum value is 5

Find the maximum and minimum values of the function y = 2-2sinx cos ^ 2x

y=2-2sinx-cos^2x
=2-2sinx-(1-sin^2x)
=2-2sinx-1+sin^2x
=1-2sinx+sin^2x
=(sinx-1)^2
The maximum value of y = 2-2sinx cos ^ 2x is 4 and the minimum value is 0