Under the given function f (x) = sinxcosx radical, 3sin2x (2 is the square of SiNx) 1. Finding the minimum positive period of F (x) 2. Find the maximum and minimum values of F (x) on the interval [0, π / 2]

f(x) = 1/2 sin2x - √3 /2 (1 - cos2x )
= 1/2 sin2x + √3/2 cos2x - √3/2
= sin(2x+ π/3) - √3/2
1. Minimum positive period T = 2 π / 2 = π
2、0 ≤ x ≤ π/2
π/3 ≤ 2x +π/3 ≤ 4π/3
When 2x + π / 3 = 4 π / 3, take the minimum value f (x) min = sin (4 π / 3) -√ 3 / 2 = - √ 3
When 2x + π / 3 = π / 2, take the maximum value f (x) max = sin (π / 2) -√ 3 / 2 = 1 - √ 3 / 2

Find the maximum value of the square + root sign 3 (sinxcosx) - 1 of the function y = SiNx, and find the x value of the maximum value Trigonometric function problems

First of all, we simplify the function expression y = 2 [- sinxsin (X-30 °) + cosxcos (X-30 °)] = 2cos (2x-30 °) and make it zero, and x = 60 ° + 90 ° * n is an integer

Let f (x) = radical 2 / [cosx (SiNx + cosx) - 1 / 2], and find the period of y = f (x)

Simplify cosx (SiNx + cosx) - 1 / 2 = cosxsinx - (cosx) ^ 2-1 / 2 = sin2x / 2 - (cosx) ^ 2-1 / 2 = sin2x / 2 - (cos2x + 1) / 2 - 1 / 2 = sin2x-cos2x = √ 2 sin (2x - Π / 4), so the function f (x) = radical 2 / [cosx (SiNx + cosx) - 1 / 2] is reduced to f (x) = √ 2 sin (2x - Π / 4)

Is the minimum positive period of the function y = SiNx radical 3 * cosx?

y=2(1/2sinx-√3/2cosx)
=2(sinxcosπ/3-cosxsinπ/3)
=2sin(x-π/3)
So t = 2 π / 1 = 2 π

Is the minimum positive period of the function y = SiNx + root 3 times cosx?

Y = SiNx + radical 3 cosx
=2 (1 / 2sinx + radical 3 / 2cosx)
=2(sinxcosπ/4+cosxsinπ/4)
=2sin(x+π/4)
The minimum positive period is 2 π

The minimum positive period of the function y = SiNx - (radical 3 cosx) is

y=2*(1/2sinx-√3/2cosx)
=2sin(x-π/3)
The minimum positive period is 2 π

Vector a = (cosx + SiNx, Radix 2 * cosx), B = (cosx SiNx, Radix 2 * SiNx), f (x) = a * B, find: (1) the minimum positive period of function f (x) Vector a = (cosx + SiNx, Radix 2 * cosx), B = (cosx SiNx, Radix 2 * SiNx), f (x) = a * B, find: (1) the minimum positive period of function f (x); (2) the monotone interval of function f (x)

The minimum positive period T = 2 π / 2 = π. The increasing interval 2K π - π / 2 ≤ 2x + π / 4 is [K π - 3 π / 8, K π + π / 8]

How to prove that the function f (x) = LG (x + root (x square + 1)) is monotone increasing function on R? Help me, thank you!

Using the definition method, the subtraction of two logarithms is equal to the division of real numbers, and then the formula of square difference is used to rationalize the molecule, and the denominator is changed into a constant, and then the numerator is compared with 0

The function y = 2 − The value range of − x2 + 4x is______ .

The definition domain should satisfy: - x2 + 4x ≥ 0, i.e. 0 ≤ x ≤ 4, y = 2 −
−x2+4x=2−
−(x−2)2+4
So when x = 2, Ymin = 0, when x = 0 or 4, ymax = 2
So the value range of the function is [0, 2],
So the answer is [0,2]

Find the value range of function y = square of root x + 1 + square of root X - 4x + 8 Find the value range of function y = (square of root x + 1) + (square of root X - 4x + 8)

You can't calculate the range of root sign
Is the square of X + 1 all under the root sign? For any x, there is the square of X + 1 > 0
Square of root X - 4x + 8 = (X-2) ^ 2 + 4 > 0
The definition domain is r

Under the given function f (x) = sinxcosx radical, 3sin2x (2 is the square of SiNx) 1. Finding the minimum positive period of F (x) 2. Find the maximum and minimum values of F (x) on the interval [0, π / 2]