The minimum positive period of the function y = 2Sin (π X / 3 + 1 / 2) For detailed explanation

The minimum positive period of the function y = 2Sin (π X / 3 + 1 / 2) For detailed explanation

T = 2 π / (π / 3) = 6. For any trigonometric function y = asin (ω x + φ) + C, the minimum positive period (or directly called period) t = 2 π / ω

What is the minimum positive period of the function y = 2Sin (π / 3 * x + 1 / 2),

T=2π/(π/3)=6
Periodic formula t = 2 π / W

The minimum positive period of the function y = √ 2Sin (x / 2 + U / 3) is

The minimum positive period of the solution function y = √ 2Sin (x / 2 + U / 3) is
T=2π/(1/2)=4π.

The minimum positive period of the function y = - 2Sin (π / 6-x / 2) The symmetry axis equation of the image with function y = sin (2x + 5 π / 2)?

T=2π/(1/2)=4π
Axis of symmetry
2x+π/2=kπ+π/2
x=kπ/2 k∈N

The minimum positive period of the function y = 2Sin (π X - 3 π)?

The minimum positive period = 2 π / π = 2, where the latter π is in front of X

(2) TaNx is the smallest periodic function of TaNx A. Pie / 2 b. pie C. Pie / 4 D.2 Pai

tanX=sinX/cosX
Bring in
Multiply cosx squared
Y = 2sinxcosx / (cosx square - SiNx Square)
=sin2X/cos2X=tan2X
A. Pie / 2

The minimum positive period of the function y = TaNx / A is A.aπ B. A π C.π/a D. π / A

The X-factor is 1 / A
The TaNx period is π
So t = π / | 1 / a | = | a | π
Choose B

Is the period of the function y = tanx-1 / TaNx?

y=tanx-1/tanx=(tanx2-1)/tanx=2(tanx2-1)/2tanx=-2(1-tanx2)/2tanx=-2/tan2x=-2cot2x
So the period is π / 2

The minimum positive period of y = 1-2tanx / 1 + TaNx The detailed process. Thank you Sorry, there is a mistake. The following should be 1 + 2tanx

If you divide the denominator of the original formula by two, you can get y = half TaNx / half + TaNx. Then according to the tangent theorem y = - Tan (x-a), the minimum positive period is π. Note: A is the angle whose tangent is half

Period of y = 1 + TaNx / 1-tanx

y=(1+tanx)/(1-tanx)=(tan45°+tanx)/(1-tan45°·tanx)=tan(x+45°)
The minimum positive period is π