Value domain of y = cos ^ 2x-1 / 2sinx + 1

Value domain of y = cos ^ 2x-1 / 2sinx + 1

Assume y = cos? X - (1 / 2) SiNx + 1
y = 1 - sin²x - (1/2)sinx + 1
= - sin²x - (1/2)sinx + 2
= -[sin²x + (1/2)sinx + 1/16 - 1/16] + 2
= -(sinx + 1/4)² + 33/16
When SiNx = - 1 / 4, the maximum value of Y is 33 / 16
When SiNx = 1, the minimum value of Y is 1 / 2
Range: [1 / 2,33 / 16]

Find the value range of y = cos ^ 2x + 2sinx-1 / 2, where x belongs to [π / 6,5 π / 6]

This simple, first simplify: y = (cosx) ^ 2 + 2sinx-1 / 2 = 1 - (SiNx) ^ 2 + 2sinx-1 / 2 = - (SiNx) ^ 2 + 2sinx + 1 / 2; because x belongs to [π / 6,5 π / 6], so SiNx belongs to [1 / 2,1]. If t = SiNx, then y = - T ^ 2 + 2T + 1 / 2; t belongs to [1 / 2,1], y = - T ^ 2 + 2T + 1 / 2

Find the value range of y = - cos (2x - π / 3) in X ∈ [0, π / 2]

Because f (x) and f (- x) are not equal, f (x) and - f (- x) are not equal. Cos (y) = cos (- y), let y = 2x - π / 3,

Find the value range of y = cos ^ 2x-4cosx + 1 x ∈ [π / 3,2 π / 3]

OK? Not y = 3cos ^ 2x-4cosx + 1

Value range y = 3 + cos (2x - π / 3)

solution
Because 2x - π / 3 ∈ R
So cos (2x - π / 3) ∈ [- 1,1]
3+cos(2x-π/3) ∈[2,4]
That is, y ∈ [2,4]

F (x) = root sign 3sinx * cos? X + 1 / 2, find the minimum positive period?

f(x)=√3sinxcos²x+(1/2)=(√3/2)sin2xcosx+(1/2)=(√3/4)(sin3x+sinx)+(1/2)；
The function f (x) is the superposition of two sinusoidal functions, and the minimum positive period is 2 π, the least common multiple of 2 π / 3 and 2 π;

Let f (x) = root sign (COS? X-sin? X) be defined as

The root is greater than 0, so cos? X is greater than sin? X, so K π - π / 4 < x < K π + π / 4 (k is an integer) draw a picture. Ask: what are k π - π / 4 and K π + π / 4? Add: are you a senior high school student, don't you know? This is the size of the angle, π stands for 180 degree ask: π is 180 ° I know, but I don't know how this k π - π / 4 comes from: cos (π / 4) = sin (π / 4) is where the absolute values of sine and cosine are equal

The maximum and minimum of the function y = cos (x + 1 / 2) ^ 2 + 1 / 4

5/4 -3/4

When the maximum value of the function f (x) = 3sinx + 2cos square x + m on the interval [0, Π / 2] is 6, find the value of constant M and the minimum value of this function when x ∈ R And find the corresponding value set of X

f(x)=3^(1/2)sinx+2(cosx)^2+m=3^(1/2)sinx+2-(sinx)^2+m
=-[(sinx)^2-3^(1/2)sinx]+(2-m)
There should be something wrong with the numbers
Otherwise, it can be transformed into the maximum value problem of quadratic function
For example, let t = SiNx,
On the interval [0, Π / 2], 0

Given that the function f (x) = 2 root sign 3sinx + 2cos square x + m, the maximum value is 2 on the interval [0, half π]. Find the value of M In the triangle ABC, if f (a) = 1 SINB = 3 sinc, the area of the triangle ABC is 3 / 4, and the root sign 3 is the length of the side a

0