What is the minimum positive period of the function y = sinxcosx?

What is the minimum positive period of the function y = sinxcosx?

y=sinxcosx=1/2sin2x
Minimum positive period T = 2 π / 2 = π

The function y = 4cos2x + 4sinxcosx-2, (x ∈ R) is known;
The known function y = 4cos2x + 4sinxcosx-2, (x ∈ R)
(1) (2) find the maximum value of the function and its corresponding x value;
(3) Write the monotone increasing interval of the function; (4) write the symmetry axis of the function

1. Y = 4cos2x + 4sinxcosx-2 = 4cos2x + 2sin2x - 2 = 2 √ 5sin (2x + T) - 2, where tank = 2, so the minimum positive period T = 2 π / 2 = π or y = 4 (cosx) ^ 2 + 4sinxcosx-2 = 2cos2x + 2 + 2sin2x-2 = 2cos2x + 2sin2x = 2 √ 2Sin (2x + π / 4) the minimum positive period T = 2 π / 2 = π if it is

The minimum positive period of function y = 2sinx ^ 2 + 3cosx ^ 2-4

y=2(1-cos2x)/2+3(1+cos2x)/2-4
=(-3+cos2x)/2
Minimum positive period T = 2 π / 2 = π

Given x ∈ R, the minimum positive period of the function f (x) = 2sinx2 + 3cosx3 is______ ．

Because the period of function y = sinx2 is 2 π 12 = 4 π, the period of function y = cosx3 is 2 π 13 = 6 π; the least common multiple of 4 π and 6 π is 12 π, so the minimum positive period of function f (x) = 2sinx2 + 3cosx3 is 12 π. So the answer is: 12 π

The range of function y = 0.3 ^| x | (x ∈ R) is

Without looking at the sign of absolute value, the function is exponential function, the base is greater than 0 and less than 1, monotonically decreases in the domain R, and the range is from 0 to positive infinity
The range of the function in the left half of the y-axis is from positive infinity to 1
The value range of the function in the right half of the Y axis is 0 to 1
Now add an absolute value to x, its range goes to the right half of the y-axis, and 0 is in the domain
So the range is (0,1]

The range of function y = x / x + 1 (x belongs to R) is?

Y = x square / x square + 1 = 1-1 / (X & # 178; + 1)
because
x²+1>=1
therefore
01-1/(1+x²)>=0
Namely
The range of the function is [0,1]
If you don't understand, you are welcome to ask,

Given that the image of the function f (x) = sin2x + mcos2x is symmetric with respect to the line x = π 8, then the coordinate of the center of symmetry of F (x) is___ ．

We know that y = sin2x + mcos2x = M2 + 1sin (2x + φ). When x = π 8, the function y = sin2x + mcos2x takes the maximum value ± M2 + 1, and substituting x = π 8, we can get: sin (2 × π 8) + MCOs (2 × π 8) = 22 (M + 1) = ± M2 + 1, and the solution is m = 1

Find the monotone increasing interval of the function f (x) = cos & sup2; (x + π / 12) + & frac12; · sin2x + 1 and draw the image of the function

F (x) = cos & sup2; (x + 12 π) + & frac12; · sin2x + 1 = (1 + cos (2x + π / 6)) / 2 + 1 / 2sin2x + 1 = 1 / 2 (COS (2x + π / 6) + sin2x) + 3 / 2 = 1 / 2 (cos2x radical 3 / 2-sin2x1 / 2 + sin2x) + 3 / 2 = 1 / 2 (cos2x * radical 3 / 2 + sin2x * 1 / 2) + 3 / 2 = radical 3 / 2Sin (2x + π / 3)

The monotone increasing interval of the function y = cos2x (- π / 4 "X" 2 π / 3) is
It means less than or equal to
I'm a science idiot

y=cos2x
2kπ

Monotone decreasing interval of function y = 1 + cos2x?
To have a detailed problem-solving process!

We know that the image of y = cosx is symmetric about the X axis and has a period of 2 π. Y = 1 + cos2x is equivalent to compressing the function laterally to one half of the original one and then translating it upward by one unit
[2K π, π + 2K π], let 2x = 2K π, then x = k π; 2x = π + 2K π, then x = π / 2 + K π. That is to say, the monotone decreasing interval of y = 1 + cos2x is [K π, π / 2 + k π]