Given function f (x) = √ 3sinxcosx cos 2 x-cos 2 + 1 / 2 (x ∈ R) (1) Finding the minimum positive period of function f (x) (2) Finding monotone interval of F (x) (3) Find the value range of the function value of F (x) on the interval (0, π, 3]

Given function f (x) = √ 3sinxcosx cos 2 x-cos 2 + 1 / 2 (x ∈ R) (1) Finding the minimum positive period of function f (x) (2) Finding monotone interval of F (x) (3) Find the value range of the function value of F (x) on the interval (0, π, 3]

∵ cos2x = 2cos? X-1  cos? X = (cos2x + 1) / 2

Vector a = (coswx + Radix 3sinwx, 1), B = (f (x), coswx), where w > 0, and a / / B, the distance between adjacent symmetry axes of the image of F (x) is 3 / 2 π

Because vector A / / vector B, we get f (x) = sin (2wx + π + 6) + 1.2. In trigonometric function, the distance between the symmetry axes adjacent to [sine. Cosine] is half a period. So the period of F (x) is 4.3 π. 2 π / W = t, so w = 3.2. So f (x) = sin (3x + 6 + π) + 1.2. That's the answer

If the symmetric axis of the image of the function y = 2Sin (x / 3 - π / 6) is a straight line, then x =? There are four options A.x=π B.x=2π C.x=3π D.x=4π

x/3-π/6=2kπ+π/2
x=6kπ+2π (k∈Z)
When k = 0, x = 2 π
Choose B
If the symmetric axis equation of y = SiNx is x = 2K π + π / 2
It's just that x in the title is x / 3 - π / 6!

Given the function y = 2Sin (3x - π / 3), X ∈ R ① The five point method is used to make a diagram of the function in a period ② This paper explains how to transform the sine curve y = SiNx to get the image of the function

If you really want to talk about images, you can
This method of a complete set of topics, counseling books are generally understood, mathematics must think of their own Oh! Come on!

The known function y = 2Sin (π / 6-1 / 3x) 1. Find its period, symmetry axis and symmetry center. 2. Find its maximum, minimum and corresponding set when taking the maximum value

y=2sin(π/6-1/3x)=-2sin（1/3x-π/6）
T = 2 π / (1 / 3) = 6 π is obtained
Let 1 / 3x - π / 6 = k π + π / 2
X = 3K π + 2 π is obtained
So the axis of symmetry of the function is x = 3K π + 2 π
Let 1 / 3x - π / 6 = k π
X = 3K π + π / 2 is obtained
So the center of symmetry of the function is (3K π + π / 2,0)
(2) Let 1 / 3x - π / 6 = 2K π + π / 2
X = 6K π + 2 π is obtained
So when x = 6K π + 2 π, the minimum value of Y is - 2
Let 1 / 3x - π / 6 = 2K π + 3 π / 2
X = 6K π + 5 π is obtained
So when x = 6K π + 5 π, the maximum value of Y is more than 2, and K is an integer

The known function y = 1 / 2Sin (3x + 6 / π) + 1 Find the value of X when (1) y is the maximum (2) Monotone decreasing interval of function and coordinates of symmetry center (3) How can the image of it be transformed from y = SiNx

(1) Let 3x + 6 / π = π / 2 + 2K π, and take K as an integer
(2) π + 2K / π, 2K / π / 2K / 2K / 2K / π / 2K / 2K / π / 2K / π / 2K / 2K / π / 2K / 2K / π / 2K / π / 2K / 2K / π / 2K / 2K / π / 2K / 2K /
3x + 6 / π = 2K π, take K as integer "
（3） X invariant y reduces 1/2 and moves up 1 unit y invariant x shifts left 6/ π and reduces to 1/3

Given the function f (x) = 2Sin (3x + y), if f (π / 3) = radical 2, then f (3 π)=

f(π/3)=2sin(π+y)=-2siny=√2
f(3π)=2sin(9π+y)=-2siny=√2

Find the function y = - cos (x 2−π 3) Monotonically increasing interval of

∵y=cos（x
2-π
3) The monotone decreasing interval of is y = - cos (x)
2-π
3) Monotonically increasing interval of,
From 2K π ≤ x
2-π
3 ≤ 2K π + π (K ∈ z), 2 π
3+4kπ≤x≤8π
3+4kπ（k∈Z），
The function y = - cos (x
2-π
3) The monotone increasing interval of the is [2 π]
3+4kπ，8π
3+4kπ]（k∈Z）．

Simplify sin (a + b) / (Sina + SINB) What is the best use of this formula

The original formula = 2Sin [(a + b) / 2] cos [(a + b) / 2] / {2Sin [(a + b) / 2] cos [(a-b) / 2]}
=cos[(a+b)/2]/cos[(a-b)/2]}

Cos (90 ° + a) cos (360 ° - a) Tan (180 ° - a) Tan (90 ° - a) / sin (270 ° + a) sin (180 ° + a)

cos(90°+a)cos(360°-a)tan(180°-a)tan(90°-a)/sin(270°+a)sin(180°+a)
=-sinacosa(-tana)cota/[(-cosa)(-sina)]
=sinacosa/(sinacosa)
=1