The minimum positive period of the function y = SiNx / 2 How to calculate it?

The minimum positive period of the function y = SiNx / 2 How to calculate it?

2π/(1/2)=4π

Find the value range of the function: (1) y = | SiNx | - 2sinx (2) y = sin | x | + | SiNx|

1、sinx≥0
y=-sinx
∴y∈[-1,0]
sinx<0
y=-3sinx
∴y∈(0,3]
2、x≥0 ①x∈[2kπ,2kπ+π] k=0,1,2. y=2sinx ∴y∈[0,2]
②x∈[2kπ+π,2kπ+2π] k=0,1,2. y=0
x<0 ①x∈[2kπ,2kπ+π] k=0,-1,-2. y=0
②x∈[2kπ+π,2kπ+2π] k=0,-1,-2. y=-2sinx y∈[-2,0]
∴y∈[-2,2]

Find the value range of function y = 2sinx-1 / SiNx + 2

y=(2sinx-1)/(sin+2)
=(2sinx+4-5)/(sinx+2)
=2-5/(sinx+2)
∵ sinx∈[-1,1]
∴ sinx+2∈[1,3]
∴ -5/(sinx+2)∈[-5,-5/3]
∴ 2-3/(cosx+2)∈[-3,1/3]
That is, the value range of the function y = 2sinx-1 / SiNx + 2 is [- 3,1 / 3]

Derivative of function y = sin squared (2x + π / 3) y'=2sin(2x+π/3)*cos(2x+π/3)*2 How does this step work out the cos (2x + π / 3) and why is it multiplied by 2

As a composite function, sin / u = 2 × u = 2
Take their derivatives and multiply them
So y '= 2Sin (2x + π / 3) * cos (2x + π / 3) * 2
y'=4sin(2x+π/3)cos(2x+π/3)=2sin(4x+2π/3)
For the question you asked, why is cos (2x + π / 3) multiplied by 2, is the derivative of V 2
Y '= 2Sin (2x + π / 3); U' = cos (2x + π / 3): V '= 2Sin (2x + π / 3) * cos (2x + π / 3) * 2

What is the primitive function of sin θ squared?

∫sin²θdθ
=(1/2)∫(1-cos2θ)dθ
=(1/2)(∫dθ-∫cos2θdθ)
=(1/2)[θ-(1/2)sin2θ]
=(1/2)θ-(1/4)sin2θ+C
If you ask ∫ sin (θ 2) d θ
This integral can't be expressed by elementary function

Find the derivative of the function y = f (x) = SiNx sin ^ 3 (x), that is, y = SiNx - (SiNx) ^ 3, or y = sinxcos ^ 2 (x) You can write the result instead of the process Find the maximum and extreme point in the (0, π / 2) interval

y'=cosx-3sin²xcosx

The minimum value of the function y = sin? X + SiNx + 2 is

y=sin^2x+sinx+2
=(sinx+1/2)^2+7/4
When SiNx = - 1 / 2, the function has a minimum value of 7 / 4

Given the function u = sin ^ 2 * x + 1 / 2 * SiNx + 1, suppose that the value of angle X is a when y is the maximum value, and B is the value of angle X when y is the minimum value Given the function y = sin ^ 2x + 1 / 2 * SiNx + 1, suppose that the value of angle X is a when y is the maximum value, and B is the value of angle X when y is the minimum value

So when SiNx = - 1 / 4, y has a minimum value, when SiNx = 1, y has a maximum value SINB = - 1 / 4, (SINB) ^ 2 + (CoSb) ^ 2 = 1, so CoSb = ± √ 15 / 4sina = 1, so cosa = 0, so sin (a-b) = sinacosb cosasinb = 1 * (± √ 15 / 4) - 0 * (- 1 / 4) = ± √ 15 / 4

It is proved that sin is not a periodic function

If it is a period, let the period be t, then sin (radical (x + T)) = sin (radical x) holds for any X
So root (x + T) = root x + 2kpi, K is an integer, PI is pi pi
If both sides are square, x + T = x + 4 (KPI) ^ 2 + 2 radical (2kpi * x) holds for any X
Therefore, t = 4 (KPI) ^ 2 + 2 root sign (2kpi * x) contradiction (because t is a fixed value, and although K on the right side can change with X, X is an arbitrary real number, and K must be an integer, which cannot guarantee that the right side is a constant, so it is contradictory)
So it's not a periodic function

It is known that f (x) is a periodic function, then f squared (x) is a periodic function? If so, how to prove it? Thank you Please make sure you prove it for me!

It is [f (x + T)] 2 = f (x) 2