The minimum positive period of y = Cotx TaNx?

The minimum positive period of y = Cotx TaNx?

y=cotx-tanx
=cosx/sinx-sinx/cosx
=(cosx)^2-(sinx)^2/sinxcosx
=2cos2x/sin2x
=2cot2x
T=π/2

What is the minimum positive period of y = TaNx?

The period is the same as y = TaNx, so the period is π

For example, y = sin (AX + b) can calculate the period according to a B. I remember there is a ω in the formula

The period of y = sin (AX + b) is 2 π / A
F (x) = asin (ω x + φ) period is t = 2 π / ω

Find the period (1) y = sin (x + π / 3) (2) y = cos2x (3) y = 3sin (x / 2 + π / 3) [please don't pay attention to the score, hope to answer my question carefully, add the correct score, thank you for your cooperation]

The minimum positive period of sine function and cosine function is t = 2 π
The minimum positive period of the function y = asin (ω x + φ): T = 2 π / ω
So the minimum positive period of (1) y = sin (x + π / 3) is 2 π
(3) The minimum positive period of y = 3sin (x / 2 + π / 3) is 2 π / 0.5 = 4 π
The minimum positive period of the function y = ACOS (ω x + φ): T = 2 π / ω
(2) The minimum positive period of y = cos2x is 2 π / 2 = π

In trigonometric functions, is sin (ω x + φ) or sin (κ ω + φ) changed periodically When sin (ω x + φ) shortens laterally and stretches κ, does the function change into sin (κ ω + φ) or sin (κ ω x + κ φ), that is, only x transformation or all independent variables transformation?

Well, the so-called elongation is the change of period. According to the periodic formula, we can see that only the coefficient change of X is given

Given that the minimum positive period of y = sin (ω x + π / 3) is π, find the value of ω Why is ω = 2 wrong, not the formula of T = 2 π / ω, ω > 0

w=±2,..
It's | 2 π / w | = π. Notice that it's an absolute value. W can be positive or negative

Find the set of independent variables X that make the following functions obtain the minimum value and maximum value, and write out what the maximum and minimum value is, y = 1-1 / 2cos1 / 3x Urgent need

y=1-1/2cos1/3x
cos(x/3)=-1
X / 3 = 2K π - π, K is an integer
x=6kπ-π,y max=3/2
cos(x/3)=1
X / 3 = 2K π, K is an integer
x=6kπ,y min=1/2
When x ∈ {x │ x = 6K π - π, K is an integer},
y max=3/2
When x ∈ {x │ x = 6K π, K is an integer},
y min=1/2

The maximum value and minimum value of the function. When the maximum value is taken, the value of X is y = 2Sin (1 / 3x + π / 3)

When 1 / 3x + π / 3 = 2K π + π / 2, i.e. x = 6K π + π / 2, sin (1 / 3x + π / 3) has a maximum value of 1, and y has a maximum value of 2
When 1 / 3x + π / 3 = 2K π - π / 2, that is, x = 6K π + 5 π / 2, sin (1 / 3x + π / 3) has a minimum value of - 1, and y has a minimum value of - 2

Find the set of independent variables X that make the following function obtain the maximum value and minimum value, and write out the maximum value and minimum value respectively. (1) y = 1-1 / 2cos Wu / 3x Write the minimum values of the following variables, x, and make the minimum values of the following functions (1) I want to adopt 3x! 2x!

From - 1 ≤ cos (π X / 3) ≤ 1, 3 / 2 ≥ 1-1 / 2 * cos (π X / 3) ≥ 1 / 2 can be obtained
When cos (π X / 3) = 1, that is, π X / 3 = 2K π, X ∈ {x | x = 6K} (K ∈ n), the minimum value of Y is 1 / 2
When cos (π X / 3) = - 1, that is, π X / 3 = 2K π + π, X ∈ {x | x = 6K + 3} (K ∈ n), the maximum value of Y is 3 / 2,

Given that f (x) = 3x ^ 2-12x + 5, when the independent variable x is in the following range, find the maximum and minimum values of the following functions （1）x∈R （2）[0,3] （3）[-1,1] Best draw a function image. Can be explained in detail

f(x)=3x^2-12x+5=3(x-2)^2-7
(1) There is no maximum, only the minimum y = - 7
(2) The minimum value y = - 7, the maximum value y = 5
(3) The minimum value y = - 4, the maximum value y = 20