F (x) = SiNx 2 prove whether it is a periodic function

F (x) = SiNx 2 prove whether it is a periodic function

f(x)=sin²x=(1-cos2x)/2=1/2-1/2cos2x
1 / 2cos2x is a periodic function
F (x) = sinx2 prove whether it is a periodic function T = 2 π / 2 = π

In the following trigonometric functions, the minimum positive period is π / 2, and the even function is
A.y=sin4x B.y=cos4x C.y=sin2x D.y=cos2x

In trigonometric function, the minimum positive period is π / 2, and the even function is
A.y=sin4x B.y=cos4x C.y=sin2x D.y=cos2x
SiNx is an odd function
The positive period of COS x is π / 2 when the period of COS x is 2 and the coefficient of COS x is 4
Choose B

The monotone increasing function of even function with π as the minimum positive period on (0, π / 4) is

That's a lot, like - cos (2x)

In the following functions, the even function which is monotonically increasing in (0, π / 2) and whose period is π is ()
A. Y = cosx b.y = | SiNx | c.1-cosx / 1 + cosx d.sin quartic X / 2-cos quartic X / 2 + 1

B, the second picture. Obviously

It is known that the function y = FX is even
If x belongs to (0, positive infinity), and FX

The solution is that the function y = FX is an even function, and it decreases when x belongs to (0, positive infinity),
Then the function y = f (x) is an increasing function when x belongs to (negative infinity, 0),
That is, when X1 and X2 belong to (negative infinity, 0) and X1 < X2, f (x1) < f (x2), and f (x1), f (x2) > 0
Then f (x) = 1 / FX
It is a decreasing function on (negative infinity, 0),
Let X1 and X2 belong to (negative infinity, 0) and X1 < x2
Then f (x1) - f (x2)
=1/f(x1)-1/f(x2)
=[f（x2）-f（x1）]/(f(x1)f(x2))
When X1 < X2, f (x1) < f (x2), that is, f (x2) - f (x1) > 0
There are also f (x1), f (x2) > 0, that is, f (x1) f (x2) > 0
That is [f (x2) - f (x1)] / (f (x1) f (x2)) > 0
That is, f (x1) - f (x2) > 0
That is, f (x1) > F (x2)
That is, f (x) = 1 / FX
It is a decreasing function on (negative infinity, 0),

If we know the function FX = root sign 1-cos square 2x, then FX is an even function with period Wu / 2. Why

f(x)=√(sin²2x)=|sin2x|
f(x+π/2)=|sin[2(x+π/2)]|
=|sin(2x+π)|
=|-sin2x|
=|sin2x|
=f(x)
So t = π / 2
f(-x)=|sin(-2x)|=|-sin2x|=|sin2x|=f(x)
And the domain of definition is r, symmetric about the origin
So it's an even function

If f (x) = x & # 178; + (M + 2) x + 1 is even function, then M =?

-2

F (x) = 1 + X & # 178 / 1-x & # 178; prove that f (x) is an even function and f (1 / x) = - f (x)

Finding f (- x) and f (1 / x) directly can be transformed into the same form as f (x)

Are the following functions odd or even: (1) y = - x; (2) y = xcosx; (3) y = x & # 178; + X

According to the definition of odd function and even function
Y = - x, y = xcosx is an odd function
Y = x & # 178; + X is a non odd non even function

Y = x & # 178; - 2 is odd function or even function, why

1) The pattern of the even function is symmetric about the Y axis in the coordinate system, and the image of the basis function is symmetric about the origin,
2) You can judge y = (- x) ^ 2-2 = x ^ 2-2 according to the definition of odd and even function, so it is even function