The period of the function y = cos2x-sin2x is______ .

The period of the function y = cos2x-sin2x is______ .

The function y = cos2x-sin2x = cos2x
∵ω=2
∴T=2π÷2=π
So the answer is: π

Why is the minimum positive period of the function y = sin ^ 4x + cos ^ 2x be 1 / 2

Y = sin ^ 4x + cos ^ 2x = sin ^ 4x + 1-sin ^ 2x = (sin ^ 2x) (sin ^ 2x-1) + 1 = (sin ^ 2x) (- cos ^ 2x) + 1 = (- 1 / 2) sin4x + 1, so the minimum positive period of function y = sin ^ 4x + cos ^ 2x t = 2 π / 4 = π / 2

What is the minimum positive period T of the function y = sin (π - x) × cos (π + x) What is the range of the function f (x) = SiNx + cosx Let y = (1 / 2) ^ X-1, (x ≤ 0) and x ^ 1 / 2 (x > 0), given f (a > 1), find the value range of real number a (1) what is the range of F (x) = SiNx + cosx (2) let the function y = (1 / 2) ^ x-1x ≤ 0) and x ^ 1 / 2 (x > 0), given f (a > 1), find the value range of real number a ("sum" represents curly bracket) (3) what is the minimum positive period T of the function y = sin (π - x) × cos (π + x)

1. SiNx + cosx = [SiNx * cos (PI / 4) + cosx * sin (PI / 4)] * sqrt2 = sqrt2 * sin (x + pi / 4); so the range is (- sqrt2, sqrt2)
There is something wrong with the topic
3. Sin (Pi-X) = SiNx, cos (PI + x) = - cosx, so y = - sinxcosx = - sin2x / 2, minimum positive period T = 2 * pi / 2 = Pi;
(PI is the legendary PI)

The value range of the function y = cos ^ 2 x-sin x is

y=cos^2 x-sin x
=1-sin^2x-sinx
=-sin^2x-sinx+1
=-[sinx+(1/2)]^2+(5/4)
∵-1≤sinx≤1
∴ymax=f(-1/2)=5/4
ymin=f(1)=1-1-1=-1
∴-1≤y≤5/4
∴y∈[-1,5/4]

The value range of the function y = sin4x + cos4x is () A. [0，1] B. [-1，1] C. [1 2，3 2] D. [1 2，1]

∵y=sin4x+cos4x
=(1−cos2x
2)2+(1+cos2x
2) 2
=2+2cos22x
Four
=1
4cos4x+3
4，
∴ymin=-1
4+3
4=1
2，ymax=1
4+3
4=1，
So the answer is: 1
2，1]
Therefore, D

Let 0 ≤ θ ≤ π / 2, function y = (1 / sin θ - 1) (1 / cos θ - 1), and find the value range of the function

 0 ≤ θ≤ π / 2,  the definition domain of the function is (0, π / 2),  0 ≤ θ / 2 ≤ π / 4,

The value range of the function y = cos (SiN x) is?

Upstairs, are we really doing the same problem
If the range of SiN x is [- 1,1], let y = cos α, then α∈ [- 1,1]
Because cos function is even function, and 1

The value range of function y = cos α - sin α can be obtained by 0 ° ≤ α < 45 °

y=√2(cosa*√2/2-sina*√2/2)
=√2(cosacos45-sinasin45)
=√2cos(a+45)
45≤a+45<90
Therefore, 00 < √ 2cos (a + 45) ≤ 1
Range (0,1]

Sin cos Tan range sin cos Tan range

Sin cos range [- 1,1]
Tan range (- infinity, + infinity)
Sin cos domain (- infinity, + infinity)
The domain of Tan is x, which cannot be equal to K π + π / 2, because the value of Tan is infinite

The value range of the function y = Tan (SiN x) is______ .

∵x∈R，∴sinx∈[-1，1]，
∵（-π
2，π
2）⊃[-1，1]，
The value range of the function y = Tan (SiN x) is [Tan (- 1), TAN1], that is [- TAN1, TAN1]
So the answer is: [- TAN1, TAN1]