Find the value range of y = cos ^ 2 x + 4sinx-2

Find the value range of y = cos ^ 2 x + 4sinx-2

Y = cos? X + 4sinx-2 = 1-sin? X + 4sinx-2 = - (sin? X-4sinx + 4) + 4-1 = - (sinx-2) 2 + 3 because: - 1 ≤ SiNx ≤ 1, the maximum value is 2 when SiNx = 1, and the minimum value is - 6 when SiNx = - 1

Y = cos (2x + π / 6) in the [0, π / 2] range is

∵x∈[0,π/2],∴2x+π/6∈[π/6,7π/6]
/ / cos (2x + π / 6) ∈ [- √ 3 / 2, √ 3 / 2] (cosine decreases in this range)
The value range is: [- √ 3 / 2, √ 3 / 2]

Value range y = 2sinx + cos ^ 2x, X ∈ [π / 6,2 π / 3)

If we know x ∈ [π / 6,2 π / 3), then: SiNx ∈ [1 / 2,1]. So when SiNx = 1, that is x = π / 2, the maximum value of the function is 2; when SiNx = 1 / 2, that is, x = π / 6, the minimum value of the function is 7 / 4

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

When x = π / 2, there is a maximum value: F (π / 2) = 3 / 2. When x = π / 6 or x = 5 π / 6, there is a minimum value

The value range of y = cos (2x + π / 3) in {0. π / 2}

If x ∈ [0, π / 2], then 2x + π / 3 ∈ [π / 3,4 π / 3], then the problem is equivalent to finding the function y = SiNx, where x ∈ [π / 3,4 π / 3], the value range of X ∈ [π / 3,4 π / 3] is obtained

The value range of the function y = cos ^ 2x + sinx-1

y=cos^2x+sinx-1
=1-sin^2x+sinx-1
=-sin^2x+sinx
=-(sinx-1/2)^2+1/4
sinx=1/2 ymax=1/4
sinx=-1 ymin=-2
The value range of function y = cos ^ 2x + sinx-1 [- 2,1 / 4]

Find the definition domain of function y = root sign (sin (x)) + root sign (- cos (x))

If there is a root sign, it means that the number in the root sign is positive
SiNx is positive in the range [2K π, 2K π + π], and cosx is negative in the range of [2K π + π / 2,2k π + 3 π / 2]
So the intersection is [2K π + π / 2,2k π + π], K belongs to the set of real numbers
You can combine the sine and cosine images

To find the maximum value of 3sinx + cos (π / 3 + x) under f (x) = radical, the detailed procedure is needed

f(x)＝√3sinx+cos(π/3+x)
＝√3sinx+cos(π/3)cosx-sin(π/3)sinx
＝√3sinx+1/2cosx-(√3/2)sinx
＝(√3/2)sinx+(1/2)cosx
＝sin(x+π/6).
When x + π / 6 = 2K π + π / 2 → x = 2K π + π / 3,
The maximum value is 1;
When x + π / 6 = 2K π + 3 π / 2 → x = 2K π + 8 π / 6,
The minimum value obtained is: - 1

Y = cos? X + root sign 3sinx + 1 x ∈ r to find the maximum value of Y

At first, the cos? X is simplified to 1-sin? X, and finally y = - sin? X + root sign 3sinx + 2 has a maximum value because - 1 < 0. According to the vertex formula of quadratic function, 11 / 4 is obtained

The minimum value of the sum of COS (COS = 2) + 1

(cosx+1/2)^2+3/4
-1