It is proved that the function y = sin (the square of x) is not a periodic function

y=sinx²=(1/2)(1-cos2x)
Its image can be obtained by cosx
1. Change the ordinate of the point on cosx image to the original half
2. Make a symmetrical image of the image obtained above with respect to the x-axis
3. Move the image obtained by 2 up one unit
4. The abscissa of the point obtained from 3. Remains unchanged and the ordinate is reduced to half of the original
As you can see, it's a periodic function. The period is π

F (x) is a periodic function. How to prove that its square is also a periodic function?

It is known that f (x) is a periodic function and its period is t
f(x+T)=f(x)
Let g (x) = f (x) ^ 2
Then G (x + T) = f (x + T) ^ 2 = f (x) ^ 2 = g (x)
Get the certificate

How to prove that y = sin (x2) is not a periodic function

prove:
Using the method of proof to the contrary
Suppose that y = sin (x ^ 2) is a periodic function and the period is t (t ≠ 0)
Then sin (x ^ 2) = sin [(x + T) ^ 2] = sin [(x-t) ^ 2]
That is: sin (x ^ 2 + 2tx + T ^ 2) = sin (x ^ 2-2tx + T ^ 2)
sin(x^2+T^2)cos(2Tx)+cos(x^2+T^2)sin(2Tx)=sin(x^2+T^2)cos(2Tx)-cos(x^2+T^2)sin(2Tx)
∴ cos(x^2+T^2)sin(2Tx)=0
However, the above formula is satisfied only when t = 0, and holds when x takes any value
contradiction
So y = sin (x ^ 2) is not a periodic function

The value range of the function y = sin squared x + sinx-2 is

y=(sinx)^2+sinx -2
=(sinx +1/2)^2 -9/4
When SiNx = - 1 / 2, y has a minimum value of - 9 / 4,
When SiNx = 1, y has a maximum value of 0,
Thus the range is [- 9 / 4,0]

The function f (x) = SiNx + sin (x + π / 2) is known, 1) Finding the minimum positive period of F (x) 2) Find the maximum and minimum of F (x) 3) If f (a) = 3 / 4, find the value of sin2a

Because f (x) = SiNx + cosx = √ 2Sin (x + π / 4) the first question t = 2 π / 1 = 2 π. The second question is that when sin (x + π / 4) = 1, it is the maximum value, that is, f (x) = √ 2Sin (x + π / 4) = - 1, that is, f (x) = - √ 2

The known function f (x) = SiNx + sin (x + π) 2），x∈R． (1) Find the minimum positive period of F (x); (2) Find the maximum and minimum values of F (x); (3) If f (α) = 3 4. Find the value of sin 2 α

(1) ∵ f (x) = SiNx + sin (π 2 + x) = SiNx + cosx = 2Sin (x + π 4) ? the minimum positive period of the function f (x) = SiN x + sin (x + π 2) is 2 π. (2) ∵ x ∈ R, - 1 ≤ SiNx ≤ 1 (2) f (x) = SiNx + sin (π 2 + x) = SiNx + cosx = 2Sin (x + π 4) ? the maximum value of F (x) is 2, the minimum

Finding the monotone interval of the function f (x) = x + SiNx

y′＝1/2+cosx
When 1 / 2 + cosx > 0,
That is 2K π - 2 π / 3

If the function f (x) = SiNx + G (x) is in the interval [- π 4，3π 4] Then the expression of function g (x) is () A. cosx B. -cosx C. 1 D. -tanx

When G (x) = cosx, the function f (x) = SiNx + G (x) = 2Sin (x + π 4), there is no monotonicity in the interval [− π 4, 3 π 4], so the function f (x) = SiNx + G (x) = 2Sin (x + π 4) is not monotonic in the interval [- π 4, 3 π 4], so option a is excluded

The function f (x) = cos2x + SiNx in the interval [- π 4，π 4] The minimum value on is () A. 2−1 Two B. -1+ Two Two C. -1 D. 1− Two Two

f（x）=cos2x+sinx=1-sin2x+sinx=-（sinx-1
2）2+5
4．
∵x∈[-π
4，π
4] So SiNx ∈ [−
Two
2，
Two
2]
Therefore, when SiNx = -
Two
When 2, the minimum value of the function is Ymin = 1 −
Two
2．
That is, if x = - π
At 4, Ymin = 1 −
Two
2．
Therefore, D

It is proved that the function y = sin (the square of x) is not a periodic function