It is known that the minimum positive period of the function f (x) = sin 2 Wx + root sign 3sinwxsin (Wx + half π) (W > 0) is π 1. Find the value of W, 2. Find the range of values on the interval [0,3 / 2 π]

f(x)=(sinwx)^2 +√3 sinwxsin(wx+ π/2)
=(sinwx)^2 +√3 sinwxcoswx
=[(1-cos2wx)/2]+(√3 /2)sin2wx
=(√3 /2)sin2wx-(1/2)cos2wx+(1/2)
=sin(2wx- π/6)+(1/2)
Then 2 π / (2W) = π
The solution w = 1
So f (x) = sin (2x - π / 6) + (1 / 2)
When 0 ≤ x ≤ 2 π / 3
0≤2x≤4π/3
-π/6≤2x- π/6≤7π/6
Then - 1 / 2 ≤ sin (2x - π / 6) ≤ 1
So the value range of F (x) is [0,3 / 2]

The known function f (x) = sin (ω x + π) 3）- 3cos（ωx+π 3) The minimum positive period of (ω > 0) is π (1) Find f (7 π) 12) Value of; (2) If △ ABC satisfies f (c) + F (B-A) = 2F (a), it is proved that △ ABC is a right triangle

(1) The function f (x) = sin (ω x + π)
3）-
3cos（ωx+π
3）=2sinωx… (2 points) (amplitude (1 point), angle 1 point,
T=2π
ω=π… (3 points), ω = 2 (4 points),
So f (7 π)
12）=2sin7π
6=-1．… (6 points)
(2) From F (c) + F (B-A) = 2F (a), sin 2C + sin (2b-2a) = 2sin2a is obtained (7 points),
-sin（2A+2B）+sin（2B-2A）=2sin2A… (8 points),
Cos2bsin2a = 0 (9 points),
So CoSb = 0 or sin2a = 0 (10 points),
Because 0 < A, B < π, so B = π
2 or a = π
2，
△abc is a right triangle (12 points)

1. Find the value range of the function y = x? - 1 / x? 2 and the value range of the function y = x + radical 1-x

A:
（1）
y=(x²-1)/(x²+1)
=(x²+1-2)/(x²+1)
=1-2/(x²+1)
Because: x? > = 0
So: x 2 + 1 > = 1,0

Another trigonometric function problem in senior one mathematics, If f (x) = Tan (x + Π), A F(O)〉F（-2）〉F（-1） B F（O）〉F（-1）〉F（-2） C F（-2〉）F（O）〉 F（-1） D F（-1）〉F（O）〉 F（-2） Why?

Choose B
Tan (x + Π) is actually TaNx moving Pai units to the left, so you just need to draw its image, the period is Pai, it is an increasing function, the feature of increasing function is that the larger x is, the greater y is, so it is b f (o) 〉 f (- 1) 〉 f (- 2)

If f (x) = sin (2x + FAI) is an even function, then a value of FAI is

f（x）=sin(2x)cos(fai)+cos(2x)sin(fai)=f（-x)= - sin(2x)cos(fai)+cos(2x)sin(fai)
be
Cos (FAI) = 0, so Fai = π / 2 + K π, a value of FAI is π / 2

Why is f (x) = sin (2x + φ) + √ 3cos (2x + φ) an even function on R?

If f (x) = sin (2x + φ) + √ 3cos (2x + φ) = 2cos (2x + φ / 6) is an even function, then φ - π / 6 = k π, K is any integer
So, φ = k π + π / 6

When f (x) = sin (2x + F), we try to find the value of F; (1) f (x) is odd function; (2) f (x) is even function

1、f(-x)=sin(-2x+F)
F (x) + F (- x) = 0
sin(2x+F)+sin(-2x+F)=0
sin2xcosF+cos2xsinF+sinFcos2x-cosFsin2x=0
2sinFcos2x=0
When sinf = 0, that is: F = k π, f (x) is an odd function!
2、f(-x)=sin(-2x+F)
f(x)-f(-x)=0
sin(2x+F)-sin(-2x+F)=0
sin2xcosF+cos2xsinF-sinFcos2x+cosFsin2x=0
2cosFsin2x=0
When cosf = 0, that is: F = k π + π / 2, f (x) is even function!

The function f (x) = sin (2x + a) is an even function

∵ f (x) = sin (2x + a) is an even function
∴f(-x)=f(x)
sin(-2x+a)=sin(2x+a)
sin(2x+a)+sin(2x-a)=0
2sin(2x)cosa=0
∵ x is an independent variable
∴cosa=0
a=kπ+π/2 (k∈Z)

If f (x) = (2x + θ) (- Pai I'm sorry. I'm sorry. It's sin (2x + θ)

The image of F (x) is a straight line, and it is an odd function when passing through the origin. So it can't be even function!

Function f (x) = sin (2x Pai / 3) (1) Find the period, maximum value of function and the corresponding value set of X (2) Finding monotone increasing interval of F (x)

T = 2 π / ω = π
f（x）max=1
∵2x-π/3=π/2+2kπ （k∈Z）
∴x=5π/12+kπ（k∈Z）
（2）；∵-π/2+2kπ≤2x-π/3≤π/2+2kπ（k∈Z）
∴-7π/12+2kπ≤x≤π/2+2kπ（k∈Z）
∴x∈（-7π/12+kπ,π/2+kπ）（k∈Z）

It is known that the minimum positive period of the function f (x) = sin 2 Wx + root sign 3sinwxsin (Wx + half π) (W > 0) is π 1. Find the value of W, 2. Find the range of values on the interval [0,3 / 2 π]