The function f (x) = sin (2x - π) 4）-2 The minimum positive period of 2sin2x is () A. π Two B. π C. 2π D. π Four

The function f (x) = sin (2x - π) 4）-2 The minimum positive period of 2sin2x is () A. π Two B. π C. 2π D. π Four

∵f（x）=sin（2x-π
4）-2
2sin2x
=
Two
2sin2x-
Two
2cos2x-
2（1-cos2x）
=
Two
2sin2x+
Two
2cos2x-
Two
=sin（2x+π
4）-
2，
The minimum positive period T = 2 π
2=π，
Therefore, B

The minimum positive period of F (x) = sin (2x - π / 4) - 2 √ 2Sin 2 x Note: 2 √ 2Sin? X, sin? X are not under the root sign

f(x) = sin(2x-π/4) - 2√2sin^2x= sin2xcosπ/4 - cos2xsinπ/4 - √2*(2sin^2x)= √2/2 sin2x - √2/2 cos2x - √2*(1-cos2x)= √2/2 sin2x - √2/2 cos2x - √2 + √2cos2x= √2/2 sin2x + √2/2 cos2x - √2= s...

The function f (x) = sin (2x + π / 6) + 2Sin? X is known 1. Find the minimum period of function f (x); 2. Find the maximum value of function f (x) and the set of values of X when the maximum value is obtained; 3. Find the monotone increasing interval of function f (x)

One
f(x)=sin(2x+π/6)+2sin²x
=sin(2x+π/6)+1-cos(2x)
=sin(2x)cos(π/6)+cos(2x)sin(π/6) -cos(2x) +1
=sin(2x)cos(π/6)+(1/2)cos(2x) -cos(2x)+1
=sin(2x)cos(π/6)-(1/2)cos(2x) +1
=sin(2x)cos(π/6)-cos(2x)sin(π/6) +1
=sin(2x-π/6) +1
Minimum positive period = 2 π / 2 = π
Two
When sin (2x - π / 6) = 1, f (x) has a maximum [f (x)] max = 1 + 1 = 2, where 2x - π / 6 = 2K π + π / 2 (K ∈ z)
x=kπ+π/3 (k∈Z)
When sin (2x - π / 6) = - 1, f (x) has a minimum value [f (x)] min = - 1 + 1 = 0, where 2x - π / 6 = 2K π - π / 2 (K ∈ z)
x=kπ-π/6 (k∈Z)
Three
When 2K π - π / 2 ≤ 2x - π / 6 ≤ 2K π + π / 2 (K ∈ z), the function increases monotonically
kπ-π/6≤x≤kπ+π/3 (k∈Z)
The monotone increasing interval of the function is [K π - π / 6, K π + π / 3] (K ∈ z)

Given that the function y = sin (Wx + a) (0 < a < π, w > 0) is an even function, then a= Thank you very much

If f (x) = sin (ω x + a) is even, then
F (x) is reduced to cosine function
∵0＜a＜π
∴a=π/2

The function y = sin (Wx + Q), (W > 0,0) is known

For even functions, x = 0 is the axis of symmetry
The axis of symmetry of sin is where the function is most worthwhile
So sin (0 * W + Q) = SINQ = 1 or - 1
Zero

1. It is known that y = 2Sin (2x + A + π / 3), if 0

Sin function is an odd function. If we want to make it even, we can change it into cos function. If a + π / 3 = k π + π / 2, we can make 2Sin (2x + A + π / 3) a cos function
The formula sin (π - a) = Sina holds in any quadrant where a is. So no matter which quadrant A is

By using the trigonometric function in the unit circle, the set of X satisfying the following conditions is found. Cos α ≥ 1 / 2

Cosine 0.5 to 1, obviously quadrant 1, 4, between plus and minus 60 degrees

Given sin θ ≤ 1 / 2 and cos θ ≤ root 3 / 2, the range of θ is determined by trigonometric function line in the unit circle

2kpi-pi/6

Using trigonometric function lines, the set of angles with cos α ≤ 1 / 2 is written

Draw the unit circle from 0 to 360 degrees,
The set of angles with cos α ≤ half is 60 to 300 degrees,
So the set of angles with cos α ≤ half is [60 + 360K, 300 + 360K]

Let a, m, n satisfy that a ^ 2 minus 4 times the root sign 2 is equal to the root sign m minus the root sign N, and find the value of positive integers a, m, n

Square the two sides of the original formula
a^2-4√2=m+n-2√(mn)
a. M and N are positive integers, and √ 2 is irrational, which can only be equivalent
m+n=a^2
√(mn)=2√2
Yes
m+n=a^2
mn=8
m. N can be 1,2,4,8
M + n Max is 9, where a max a = 3
A can be taken as 1,2,3
A = 1 impossible,
When Mn = 4, Mn = 4,
m. N is regarded as the solution of the equation x ^ 2-4x + 8 = 0
If the discriminant is less than 0, there is no solution
So a can only be 3, m, n. One of them is 1 and the other is 8
The left side of the original equation is √ 81 - √ 32 > 0, and the right side is greater than 0, M > N, M = 8, n = 1
So,
a=3,m=8,n=1
You have hope,