Find the maximum value of the function f (x) = cosx cosx square + 1 / 2

Find the maximum value of the function f (x) = cosx cosx square + 1 / 2

f(x)=cosx-(cosx)^2+1/2
=-(cosx-1/2)^2+3/4
Because - 1 ≤ cosx ≤ 1
So - 3 / 2 ≤ cosx-1 / 2 ≤ 1 / 2
So 0 ≤ (cosx-1 / 2) ^ 2 ≤ 9 / 4
Then - 3 / 2 ≤ f (x) ≤ 3 / 4
So the maximum is 3 / 4
If you don't understand, please hi me, I wish you a happy study!

The period of function f (x) = cosx's Square-1 / 2

f(x)=(1+cos2x)/2-1/2
=(cos2x)/2
T=2π/2=π

The maximum of the square of the function y = SiNx + cosx

y=sinx+cos²x=sinx+1-sin²x=-sin²x+sinx+1=-（sinx-1/2）²+5/4,
So when SiNx = 1 / 2, the function y = SiNx + cos & # 178; X has a maximum value of 5 / 4~

Find the square of SiNx times the maximum of cosx?

According to the meaning of the question, y = (SiNx) ^ 2 * cosx, because: (SiNx) ^ 2 + (cosx) ^ 2 = 1 means:
Y = [1 - (cosx) ^ 2] cosx = cosx - (cosx) ^ 3
Y '= - SiNx + 3 (cosx) ^ 2 * SiNx, let y' = 0, have: - SiNx + 3 (cosx) ^ 2 * SiNx = 0
( cosx )^2=1/3 cosx=(1/3) ^(1/2)=3^(1/2)/3.
(sinx)^2=1-( cosx )^2=1-1/3=2/3
Comparing the value of the function when cosx = 3 ^ (1 / 2) / 3, we know that when cosx = 3 ^ (1 / 2) / 3, we can
Take the maximum value, y (max) = (2 / 3) * 3 ^ (1 / 2) / 3 = 2 * 3 ^ (1 / 2) / 9
So: the maximum value of the square of SiNx multiplied by cosx is 2 * 3 ^ (1 / 2) / 9

Find the maximum value of SiNx * cosx * cosx between 0 and Pai / 2
The derivative is cosx ^ 3-2sinx * SiNx * cosx. It should be right, but there is still no way

Original formula = SiNx * (cosx) ^ 2 = SiNx * (1 - (SiNx) ^ 2)
Let SiNx = t (0 = < T)

g(x)=cos(sinx),0

0

Let g (x) = cos (SiNx), (0 ≤ x ≤ π) find the maximum and minimum of G (x)
Process

0 ≤ x ≤ π, the range of SiN x is [0,1]
The range of COS (SiN x) is [cos 1,1] (taking SiN x as a whole)

When x ∈ [- Π / 4, Π / 4], f '(x) = cosx-1 / cos ^ x = [(cosx) ^ 3-1) / (cosx) ^ 2

Well, No
f(x)=sinx-tanx
The derivative f '(x) = cosx-1 / (cosx) ^ 2 of F (x)
If the derivative is less than 0, the original function decreases monotonously

Given the function f (x) = cosx + cos (x + π / 3). (1) find the monotone decreasing interval of the function on [0, π]
(1) Find the monotone decreasing interval of the function on [0, π]
(2) If x is an acute angle, find the range of the function

F (x) = cosx + cos (x + π / 3) = 3 / 2cosx - √ 3 / 2sinx = √ 3 (√ 3 / 2cosx-1 / 2sinx) = √ 3cos (x + π / 6) x ∈ [0, π] x + π / 6 ∈ [π / 6,7 π / 6] [0,5 π / 6] is a monotone decreasing interval [5 π / 6, π] is a monotone increasing interval, X is an acute angle, X ∈ (0, π / 2) x + π / 6 ∈ (π / 6,2

The range of F (x) = (1-sin2x) / cosx is known···
The known function f (x) = (1-sin2x) / cosx
Maybe I'm too busy··

f（x）=（1-sin2x）/cosx
=(sinx-cosx)^2/cosx
=(1-tanx)^2
If TaNx is a real number, then f (x) is a real number