Let the minimum value of the function y = - 2Sin square x-2acosx-2a + 1 on X is f (a). Try to determine that f (a) 0 = 1 / 2 is satisfied and find the maximum value of Y for the value of a at this time

Let the minimum value of the function y = - 2Sin square x-2acosx-2a + 1 on X is f (a). Try to determine that f (a) 0 = 1 / 2 is satisfied and find the maximum value of Y for the value of a at this time

If f (x) = - 2Sin 2 x-2acosx-2a + 1F (x) = 2cos 2 x-2acos-2a-1f (x) = 2 × [cosx - (A / 2)] 2 - [(1 / 2) a} + 2A + 1] function f (x) is g (a), then: {f (- 1) = 1 (A2) if G (a) = 1 / 2, then: (1) if - 2 ≤ a ≤ 2, then: - (1 / 2) a

If the minimum value of the function f (x) = cos2x-2acosx + A ^ 2-2a (0 ≤ x ≤ π / 2) is - 2 And find the maximum value of F (x) The process should be detailed

F (x) = cos2x-2acosx + A ^ 2-2a = 2cos ^ 2 x-1-2acosx + A ^ 2-2a = 2 [cos ^ 2 x - (1 / 2) a] ^ 2 + (1 / 2) a ^ 2-2a-1 when cosx = (1 / 2) a, f (x) has a minimum value of (1 / 2) a ^ 2-2a-1; from the meaning of the title: (1 / 2) a ^ 2-2a-1 = - 2 (1) And 0

If the minimum value of the function f (x) = cos2x-2acosx + A ^ 2-2a (0 ≤ x ≤ π / 2) is - 2, then the value of a is

The minimum value of F (x) = cos2x-2acosx + A ^ 2-2a = (a-cosx-1) ^ 2 + (cosx-1) ^ 2-2 is - 2
Then there is: (a-cosx-1) ^ 2 + (cosx-1) ^ 2 = 0
Cosx = 1; a = 2

Function y = cos2x-2acosx-2a (1) find the minimum value of F (a) (2) try to determine the value of a when f (a) = 1 / 2 and find the maximum value of Y at this time Please,

(1)f(a)=cos2a-2acosa-2a=2(cosa)^2-2acosa-2a-1
If t = cosa, then - 1=

Given that the minimum value of the function f (x) = cos2x-2acosx + 2 in the interval (0, π) is g (a), the analytic formula of G (a) is obtained, and the function y = g (a) is pointed out

F (x) = cos2x-2acosx + 2 = 2 * (cosx) ^ 2-2acosx + 1 = 2 (cosx-a / 2) ^ 2-A ^ 2 / 2 + 1cosx on the interval (- 1,1), when a / 2 = - 1, the function has no minimum value in the interval, because the minimum value of x = A / 2 is g (a) = 1-A ^ 2 / 2

On the function of X, y = cos2x-2acosx-2a < 1 > find the minimum value of F (a) < 2 > try to determine the value satisfying f (a) = 1 / 2a and get the maximum value of Y at this time Please,

If y = cos2x-2acosx-2a = 2cos? X-1-2acosx-2a = 2cos? X-2acosx-1-2a, let cosx = t - 1 ≤ t ≤ T, then y = 2T? - 2at-1-2a = 2 (T-A / 2) - 1-2a-a? / 2 when a / 2 ＜ - 1, the minimum value is: F (a) = 1 when - 1 ≤ A / 2 ≤ 1, the minimum value is: F (a) = - A / 2 -

If the minimum value of the function y = 2cosx + B is - 3, find the maximum value of the function. 2) find the minimum value of the function y = sin? X-cos? X

The minimum value of the function y = 2cosx + B is - 3
2*(-1)+b=-3
b=-1
Maximum = 2 * 1-1 = 1
y=sin²x-cos²x
=sin²x-(1-sin²x)
=2sin²x-1
When SiNx = 0, the function y has a minimum value = - 1

The function f (x) = cosx-2cosx * sin 2 (α / 2) - sinxsin α (0

Solution: simplify f (x): F (x) = (cosx) * [1-2sin ^ 2 (α / 2)] - sinxsin α = (cosx) * cos α - sinxsin α (double angle formula) = cos (x + α) (cosine angle sum formula) (1) since f (x) has a minimum value when x = π / 2, then f (π / 2) = - 1, that is, π / 2 + α = 2K π + π (k is an integral number), α = 2K π + π / 2 is obtained

When x ∈ [π / 3,4 π / 3], the function y = sin? 2x + 2cosx has the maximum and minimum value

y=sin²x+2cosx
=1-cos²x+2cosx
=2-(cosx-1)²
x∈[π/3,4π/3]
-1≤cosx≤1/2
When cosx = 1 / 2, the maximum value is 2-1 / 4 = 7 / 4;
When cosx = - 1, the minimum value is 2-4 = - 2

Given the function f (x) = sin? 2x + 2cosx, find the maximum value and the corresponding x value when 0 ≤ x ≤ π / 2

Because sin? X + cos? X = 1
So sin? X = 1-cos? X
y=f(x)=-cos²x+2cosx+1
Let cosx = t, because 0 ≤ x ≤ π / 2, 0 ≤ cosx ≤ 1, that is, 0 ≤ t ≤ 1;
y=-t²+2t+1
The symmetry axis of quadratic parabola with opening downward is t = 1;
So it is increasing in the interval 0 ≤ t ≤ 1,
So when the maximum value is t = 1, y = 2;
Cosx = 1, x = 0;